S857770101

AHS8577701

FOR OFFICIAL USE

Fill in these boxes and read what is printed below

Number of seat

Town

copy

Mark

Full name of centre

Forename(s) Surname

Scottish candidate numberDate of birth

YearDay Month

NationalQualicationsSPECIMEN ONLY

Total marks mdash 155

Attempt ALL questions

Reference may be made to the Physics Relationships Sheet S8577711 and the Data Sheet on page 02

Write your answers clearly in the spaces provided in this booklet Additional space for answers and rough work is provided at the end of this booklet If you use this space you must clearly identify the question number you are attempting Any rough work must be written in this booklet You should score through your rough work when you have written your final copy

Care should be taken to give an appropriate number of significant figures in the final answers to calculations

Use blue or black ink

Before leaving the examination room you must give this booklet to the Invigilator if you do not you may lose all the marks for this paper

S8577701

Date mdash Not applicable

Duration mdash 3 hours

Physics

S857770102page 02

DATA SHEETCOMMON PHYSICAL QUANTITIES

Quantity Symbol Value Quantity Symbol Value

Gravitational acceleration on EarthRadius of EarthMass of EarthMass of JupiterRadius of JupiterMean Radius of Jupiter OrbitSolar radiusMass of Sun1 AUStefan-Boltzmann constantUniversal constant of gravitation

gREMEMJRJ

σ

G

9middot8 m sminus2

6middot4 times 106 m6middot0 times 1024 kg1middot90 times 1027 kg7middot15 times 107 m

7middot79 times 1011 m6middot955 times 108 m2middot0 times 1030 kg1middot5 times 1011 m

5middot67 times 10minus8 W mminus2 Kminus4

6middot67 times 10minus11 m3 kgminus1 sminus2

Mass of electronCharge on electronMass of neutronMass of protonMass of alpha particleCharge on alpha particlePlanckrsquos constantPermittivity of free spacePermeability of free spaceSpeed of light in vacuumSpeed of sound in air

meemnmpmα

h

ε0

μ0

c

v

9middot11 times 10minus31 kgminus1middot60 times 10minus19 C1middot675 times 10minus27 kg1middot673 times 10minus27 kg6middot645 times 10minus27 kg

3middot20 times 10minus19 C6middot63 times 10minus34 J s

8middot85 times 10minus12 F mminus1

4π times 10minus7 H mminus1

3middot00 times 108 m sminus1

3middot4 times 102 m sminus1

REFRACTIVE INDICESThe refractive indices refer to sodium light of wavelength 589 nm and to substances at a temperature of 273 K

Substance Refractive index Substance Refractive index

DiamondGlassIcePerspex

2middot421middot51 1middot311middot49

GlycerolWaterAirMagnesium Fluoride

1middot47 1middot331middot001middot38

SPECTRAL LINES

Element Wavelength (nm) Colour Element Wavelength (nm) Colour

Hydrogen

Sodium

656486434410397389

589

RedBlue-greenBlue-violetVioletUltravioletUltraviolet

Yellow

Cadmium 644509480

RedGreenBlue

LasersElement Wavelength (nm) Colour

Carbon dioxide

Helium-neon

955010 590

633

Infrared

Red

PROPERTIES OF SELECTED MATERIALS

Substance Density(kg mminus3)

MeltingPoint(K)

BoilingPoint(K)

Specific HeatCapacity

(J kgminus1 Kminus1)

Specific LatentHeat ofFusion(J kgminus1)

Specific LatentHeat of

Vaporisation(J kgminus1)

AluminiumCopperGlassIceGlycerolMethanolSea WaterWaterAirHydrogenNitrogenOxygen

2middot70 times 103

8middot96 times 103

2middot60 times 103

9middot20 times 102

1middot26 times 103

7middot91 times 102

1middot02 times 103

1middot00 times 103

1middot299middot0 times 10minus2

1middot251middot43

93313571400273291175264273

146355

26232853 563338377373

207790

9middot02 times 102

3middot86 times 102

6middot70 times 102

2middot10 times 103

2middot43 times 103

2middot52 times 103

3middot93 times 103

4middot18 times 103

1middot43 times 104

1middot04 times 103

9middot18 times 102

3middot95 times 105

2middot05 times 105

3middot34 times 105

1middot81 times 105

9middot9 times 104

3middot34 times 105

8middot30 times 105

1middot12 times 106

2middot26 times 106

4middot50 times 105

2middot00 times 105

2middot40 times 104

The gas densities refer to a temperature of 273 K and a pressure of 1middot01 times 105 Pa

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Total marks mdash 155 marks

Attempt ALL questions

1

A car on a long straight track accelerates from rest The carrsquos run begins at time t = 0

Its velocity v at time t is given by the equation20middot135 1middot26v t t= +

where v is measured in m sminus1

and t is measured in s

Using calculus methods

(a) determine the acceleration of the car at t = 15middot0 s

Space for working and answer

(b) determine the displacement of the car from its original position at this time

Space for working and answer

3

3

[Turn over

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2 Water is removed from clothes during the spin cycle of a washing machine The drum holding the clothes has a maximum spin rate of 1250 revolutions per minute

Figure 2A

(a) Show that the maximum angular velocity of the drum is 131 rad sminus1

Space for working and answer

(b) At the start of a spin cycle the drum has an angular velocity of 7∙50 rad sminus1 It then takes 12∙0 seconds to accelerate to the maximum angular velocity

    (i)  Calculate the angular acceleration of the drum during the 12∙0 seconds assuming the acceleration is uniform

Space for working and answer

2

3

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2 (b) (continued)

(ii) Determine how many revolutions the drum will make during the 12∙0 seconds

Space for working and answer

(c) When the drum is rotating at maximum angular velocity an item of wet clothing of mass 1∙5 times 10minus2 kg rotates at a distance of 0∙28 m from the axis of rotation as shown in Figure 2B

Figure 2B

0∙28 m

item of clothing

Calculate the centripetal force acting on the item of clothing

Space for working and answer

4

3

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2 (continued)

(d) The outer surface of the drum has small holes as shown in Figure 2C These holes allow most of the water to be removed

Figure 2C

(i) Explain why the water separates from the item of clothing during the spin cycle

(ii) The drum rotates in an anticlockwise direction

Indicate on Figure 2D the direction taken by a water droplet as it leaves the drum

Figure 2D

water droplet

drum

2

1

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2 (d) (continued)

(iii) Explain what happens to the magnitude of the force acting on an item of clothing inside the drum as it rotates at its maximum angular velocity

[Turn over

2

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3 A disc of mass 6∙0 kg and radius 0∙50 m is allowed to rotate freely about its central axis as shown in Figure 3A

central axis

Figure 3A

(a) Show that the moment of inertia of the disc is 0∙75 kg m2

Space for working and answer

(b) The disc is rotating with an angular velocity of 12 rad sminus1 A cube of mass 2∙0 kg is then dropped onto the disc The cube remains at a distance of 0∙40 m from the axis of rotation as shown in Figure 3B

central axis

040 m

Figure 3B

2

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3 (b) (continued)

(i) Determine the total moment of inertia of the disc and cube

Space for working and answer

(ii) Calculate the angular velocity of the disc after the cube lands

Space for working and answer

(iii) State one assumption you have made in your response to b(ii)

[Turn over

3

3

1

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3 (continued)

(c) The cube is removed and the disc is again made to rotate with a constant angular velocity of 12 rad sminus1 A sphere of mass 2∙0 kg is then dropped onto the disc at a distance of 0∙40 m from the axis as shown in Figure 3C

040 m

Figure 3C

State whether the resulting angular velocity of the disc is greater than the same as or less than the value calculated in b(ii)

Justify your answer 2

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4 The International Space Station (ISS) is in orbit around the Earth

ISS

40 times 105 m

not to scale

Figure 4A

(a) (i) The gravitational pull of the Earth keeps the ISS in orbit

Show that for an orbit of radius r the period T is given by the expression

3

2E

rT πGM

=

where the symbols have their usual meaning

(ii) Calculate the period of orbit of the ISS when it is at an altitude of 4∙0 times 105 m above the surface of the Earth

Space for working and answer

2

2

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4 (continued)

(b) The graph in Figure 4B shows how the altitude of the ISS has varied over time Reductions in altitude are due to the drag of the Earthrsquos atmosphere acting on the ISS

alti

tude

(times10

5 m

)

418

417

416

415

414

413

412

01 N

ov 2

013

01 D

ec 2

013

31 D

ec 2

013

30 J

an 2

014

01 M

ar 2

014

31 M

ar 2

014

30 A

pr 2

014

30 M

ay 2

014

29 J

un 2

014

29 J

ul 2

014

28 A

ug 2

014

27 S

ep 2

014

27 O

ct 2

014

26 N

ov 2

014

date

Figure 4B

(i) Determine the value of Earthrsquos gravitational field strength at the ISS on 1 March 2014

Space for working and answer

4

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4 (b) (continued)

(ii) In 2011 the average altitude of the ISS was increased from 350 km to 400 km

Suggest an advantage of operating the ISS at this higher altitude

(c) Clocks designed to operate on the ISS are synchronised with clocks on Earth before they go into space On the ISS a correction factor is necessary for the clocks to remain synchronised with clocks on Earth

Explain why this correction factor is necessary

[Turn over

1

2

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 ESP 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 JPN 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S857770102page 02

DATA SHEETCOMMON PHYSICAL QUANTITIES

Quantity Symbol Value Quantity Symbol Value

Gravitational acceleration on EarthRadius of EarthMass of EarthMass of JupiterRadius of JupiterMean Radius of Jupiter OrbitSolar radiusMass of Sun1 AUStefan-Boltzmann constantUniversal constant of gravitation

gREMEMJRJ

σ

G

9middot8 m sminus2

6middot4 times 106 m6middot0 times 1024 kg1middot90 times 1027 kg7middot15 times 107 m

7middot79 times 1011 m6middot955 times 108 m2middot0 times 1030 kg1middot5 times 1011 m

5middot67 times 10minus8 W mminus2 Kminus4

6middot67 times 10minus11 m3 kgminus1 sminus2

Mass of electronCharge on electronMass of neutronMass of protonMass of alpha particleCharge on alpha particlePlanckrsquos constantPermittivity of free spacePermeability of free spaceSpeed of light in vacuumSpeed of sound in air

meemnmpmα

h

ε0

μ0

c

v

9middot11 times 10minus31 kgminus1middot60 times 10minus19 C1middot675 times 10minus27 kg1middot673 times 10minus27 kg6middot645 times 10minus27 kg

3middot20 times 10minus19 C6middot63 times 10minus34 J s

8middot85 times 10minus12 F mminus1

4π times 10minus7 H mminus1

3middot00 times 108 m sminus1

3middot4 times 102 m sminus1

REFRACTIVE INDICESThe refractive indices refer to sodium light of wavelength 589 nm and to substances at a temperature of 273 K

Substance Refractive index Substance Refractive index

DiamondGlassIcePerspex

2middot421middot51 1middot311middot49

GlycerolWaterAirMagnesium Fluoride

1middot47 1middot331middot001middot38

SPECTRAL LINES

Element Wavelength (nm) Colour Element Wavelength (nm) Colour

Hydrogen

Sodium

656486434410397389

589

RedBlue-greenBlue-violetVioletUltravioletUltraviolet

Yellow

Cadmium 644509480

RedGreenBlue

LasersElement Wavelength (nm) Colour

Carbon dioxide

Helium-neon

955010 590

633

Infrared

Red

PROPERTIES OF SELECTED MATERIALS

Substance Density(kg mminus3)

MeltingPoint(K)

BoilingPoint(K)

Specific HeatCapacity

(J kgminus1 Kminus1)

Specific LatentHeat ofFusion(J kgminus1)

Specific LatentHeat of

Vaporisation(J kgminus1)

AluminiumCopperGlassIceGlycerolMethanolSea WaterWaterAirHydrogenNitrogenOxygen

2middot70 times 103

8middot96 times 103

2middot60 times 103

9middot20 times 102

1middot26 times 103

7middot91 times 102

1middot02 times 103

1middot00 times 103

1middot299middot0 times 10minus2

1middot251middot43

93313571400273291175264273

146355

26232853 563338377373

207790

9middot02 times 102

3middot86 times 102

6middot70 times 102

2middot10 times 103

2middot43 times 103

2middot52 times 103

3middot93 times 103

4middot18 times 103

1middot43 times 104

1middot04 times 103

9middot18 times 102

3middot95 times 105

2middot05 times 105

3middot34 times 105

1middot81 times 105

9middot9 times 104

3middot34 times 105

8middot30 times 105

1middot12 times 106

2middot26 times 106

4middot50 times 105

2middot00 times 105

2middot40 times 104

The gas densities refer to a temperature of 273 K and a pressure of 1middot01 times 105 Pa

S857770103page 03

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Total marks mdash 155 marks

Attempt ALL questions

1

A car on a long straight track accelerates from rest The carrsquos run begins at time t = 0

Its velocity v at time t is given by the equation20middot135 1middot26v t t= +

where v is measured in m sminus1

and t is measured in s

Using calculus methods

(a) determine the acceleration of the car at t = 15middot0 s

Space for working and answer

(b) determine the displacement of the car from its original position at this time

Space for working and answer

3

3

[Turn over

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2 Water is removed from clothes during the spin cycle of a washing machine The drum holding the clothes has a maximum spin rate of 1250 revolutions per minute

Figure 2A

(a) Show that the maximum angular velocity of the drum is 131 rad sminus1

Space for working and answer

(b) At the start of a spin cycle the drum has an angular velocity of 7∙50 rad sminus1 It then takes 12∙0 seconds to accelerate to the maximum angular velocity

    (i)  Calculate the angular acceleration of the drum during the 12∙0 seconds assuming the acceleration is uniform

Space for working and answer

2

3

S857770105page 05

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2 (b) (continued)

(ii) Determine how many revolutions the drum will make during the 12∙0 seconds

Space for working and answer

(c) When the drum is rotating at maximum angular velocity an item of wet clothing of mass 1∙5 times 10minus2 kg rotates at a distance of 0∙28 m from the axis of rotation as shown in Figure 2B

Figure 2B

0∙28 m

item of clothing

Calculate the centripetal force acting on the item of clothing

Space for working and answer

4

3

[Turn over

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2 (continued)

(d) The outer surface of the drum has small holes as shown in Figure 2C These holes allow most of the water to be removed

Figure 2C

(i) Explain why the water separates from the item of clothing during the spin cycle

(ii) The drum rotates in an anticlockwise direction

Indicate on Figure 2D the direction taken by a water droplet as it leaves the drum

Figure 2D

water droplet

drum

2

1

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2 (d) (continued)

(iii) Explain what happens to the magnitude of the force acting on an item of clothing inside the drum as it rotates at its maximum angular velocity

[Turn over

2

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3 A disc of mass 6∙0 kg and radius 0∙50 m is allowed to rotate freely about its central axis as shown in Figure 3A

central axis

Figure 3A

(a) Show that the moment of inertia of the disc is 0∙75 kg m2

Space for working and answer

(b) The disc is rotating with an angular velocity of 12 rad sminus1 A cube of mass 2∙0 kg is then dropped onto the disc The cube remains at a distance of 0∙40 m from the axis of rotation as shown in Figure 3B

central axis

040 m

Figure 3B

2

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3 (b) (continued)

(i) Determine the total moment of inertia of the disc and cube

Space for working and answer

(ii) Calculate the angular velocity of the disc after the cube lands

Space for working and answer

(iii) State one assumption you have made in your response to b(ii)

[Turn over

3

3

1

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3 (continued)

(c) The cube is removed and the disc is again made to rotate with a constant angular velocity of 12 rad sminus1 A sphere of mass 2∙0 kg is then dropped onto the disc at a distance of 0∙40 m from the axis as shown in Figure 3C

040 m

Figure 3C

State whether the resulting angular velocity of the disc is greater than the same as or less than the value calculated in b(ii)

Justify your answer 2

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4 The International Space Station (ISS) is in orbit around the Earth

ISS

40 times 105 m

not to scale

Figure 4A

(a) (i) The gravitational pull of the Earth keeps the ISS in orbit

Show that for an orbit of radius r the period T is given by the expression

3

2E

rT πGM

=

where the symbols have their usual meaning

(ii) Calculate the period of orbit of the ISS when it is at an altitude of 4∙0 times 105 m above the surface of the Earth

Space for working and answer

2

2

[Turn over

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4 (continued)

(b) The graph in Figure 4B shows how the altitude of the ISS has varied over time Reductions in altitude are due to the drag of the Earthrsquos atmosphere acting on the ISS

alti

tude

(times10

5 m

)

418

417

416

415

414

413

412

01 N

ov 2

013

01 D

ec 2

013

31 D

ec 2

013

30 J

an 2

014

01 M

ar 2

014

31 M

ar 2

014

30 A

pr 2

014

30 M

ay 2

014

29 J

un 2

014

29 J

ul 2

014

28 A

ug 2

014

27 S

ep 2

014

27 O

ct 2

014

26 N

ov 2

014

date

Figure 4B

(i) Determine the value of Earthrsquos gravitational field strength at the ISS on 1 March 2014

Space for working and answer

4

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4 (b) (continued)

(ii) In 2011 the average altitude of the ISS was increased from 350 km to 400 km

Suggest an advantage of operating the ISS at this higher altitude

(c) Clocks designed to operate on the ISS are synchronised with clocks on Earth before they go into space On the ISS a correction factor is necessary for the clocks to remain synchronised with clocks on Earth

Explain why this correction factor is necessary

[Turn over

1

2

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

S857770115page 15

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

[Turn over

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

[Turn over

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

[Turn over

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

[Turn over

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

[Turn over

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

[Turn over

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

S857770129page 29

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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S857770103page 03

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Total marks mdash 155 marks

Attempt ALL questions

1

A car on a long straight track accelerates from rest The carrsquos run begins at time t = 0

Its velocity v at time t is given by the equation20middot135 1middot26v t t= +

where v is measured in m sminus1

and t is measured in s

Using calculus methods

(a) determine the acceleration of the car at t = 15middot0 s

Space for working and answer

(b) determine the displacement of the car from its original position at this time

Space for working and answer

3

3

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2 Water is removed from clothes during the spin cycle of a washing machine The drum holding the clothes has a maximum spin rate of 1250 revolutions per minute

Figure 2A

(a) Show that the maximum angular velocity of the drum is 131 rad sminus1

Space for working and answer

(b) At the start of a spin cycle the drum has an angular velocity of 7∙50 rad sminus1 It then takes 12∙0 seconds to accelerate to the maximum angular velocity

    (i)  Calculate the angular acceleration of the drum during the 12∙0 seconds assuming the acceleration is uniform

Space for working and answer

2

3

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2 (b) (continued)

(ii) Determine how many revolutions the drum will make during the 12∙0 seconds

Space for working and answer

(c) When the drum is rotating at maximum angular velocity an item of wet clothing of mass 1∙5 times 10minus2 kg rotates at a distance of 0∙28 m from the axis of rotation as shown in Figure 2B

Figure 2B

0∙28 m

item of clothing

Calculate the centripetal force acting on the item of clothing

Space for working and answer

4

3

[Turn over

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2 (continued)

(d) The outer surface of the drum has small holes as shown in Figure 2C These holes allow most of the water to be removed

Figure 2C

(i) Explain why the water separates from the item of clothing during the spin cycle

(ii) The drum rotates in an anticlockwise direction

Indicate on Figure 2D the direction taken by a water droplet as it leaves the drum

Figure 2D

water droplet

drum

2

1

S857770107page 07

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2 (d) (continued)

(iii) Explain what happens to the magnitude of the force acting on an item of clothing inside the drum as it rotates at its maximum angular velocity

[Turn over

2

S857770108page 08

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3 A disc of mass 6∙0 kg and radius 0∙50 m is allowed to rotate freely about its central axis as shown in Figure 3A

central axis

Figure 3A

(a) Show that the moment of inertia of the disc is 0∙75 kg m2

Space for working and answer

(b) The disc is rotating with an angular velocity of 12 rad sminus1 A cube of mass 2∙0 kg is then dropped onto the disc The cube remains at a distance of 0∙40 m from the axis of rotation as shown in Figure 3B

central axis

040 m

Figure 3B

2

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3 (b) (continued)

(i) Determine the total moment of inertia of the disc and cube

Space for working and answer

(ii) Calculate the angular velocity of the disc after the cube lands

Space for working and answer

(iii) State one assumption you have made in your response to b(ii)

[Turn over

3

3

1

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3 (continued)

(c) The cube is removed and the disc is again made to rotate with a constant angular velocity of 12 rad sminus1 A sphere of mass 2∙0 kg is then dropped onto the disc at a distance of 0∙40 m from the axis as shown in Figure 3C

040 m

Figure 3C

State whether the resulting angular velocity of the disc is greater than the same as or less than the value calculated in b(ii)

Justify your answer 2

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4 The International Space Station (ISS) is in orbit around the Earth

ISS

40 times 105 m

not to scale

Figure 4A

(a) (i) The gravitational pull of the Earth keeps the ISS in orbit

Show that for an orbit of radius r the period T is given by the expression

3

2E

rT πGM

=

where the symbols have their usual meaning

(ii) Calculate the period of orbit of the ISS when it is at an altitude of 4∙0 times 105 m above the surface of the Earth

Space for working and answer

2

2

[Turn over

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4 (continued)

(b) The graph in Figure 4B shows how the altitude of the ISS has varied over time Reductions in altitude are due to the drag of the Earthrsquos atmosphere acting on the ISS

alti

tude

(times10

5 m

)

418

417

416

415

414

413

412

01 N

ov 2

013

01 D

ec 2

013

31 D

ec 2

013

30 J

an 2

014

01 M

ar 2

014

31 M

ar 2

014

30 A

pr 2

014

30 M

ay 2

014

29 J

un 2

014

29 J

ul 2

014

28 A

ug 2

014

27 S

ep 2

014

27 O

ct 2

014

26 N

ov 2

014

date

Figure 4B

(i) Determine the value of Earthrsquos gravitational field strength at the ISS on 1 March 2014

Space for working and answer

4

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4 (b) (continued)

(ii) In 2011 the average altitude of the ISS was increased from 350 km to 400 km

Suggest an advantage of operating the ISS at this higher altitude

(c) Clocks designed to operate on the ISS are synchronised with clocks on Earth before they go into space On the ISS a correction factor is necessary for the clocks to remain synchronised with clocks on Earth

Explain why this correction factor is necessary

[Turn over

1

2

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

[Turn over

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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2 Water is removed from clothes during the spin cycle of a washing machine The drum holding the clothes has a maximum spin rate of 1250 revolutions per minute

Figure 2A

(a) Show that the maximum angular velocity of the drum is 131 rad sminus1

Space for working and answer

(b) At the start of a spin cycle the drum has an angular velocity of 7∙50 rad sminus1 It then takes 12∙0 seconds to accelerate to the maximum angular velocity

    (i)  Calculate the angular acceleration of the drum during the 12∙0 seconds assuming the acceleration is uniform

Space for working and answer

2

3

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2 (b) (continued)

(ii) Determine how many revolutions the drum will make during the 12∙0 seconds

Space for working and answer

(c) When the drum is rotating at maximum angular velocity an item of wet clothing of mass 1∙5 times 10minus2 kg rotates at a distance of 0∙28 m from the axis of rotation as shown in Figure 2B

Figure 2B

0∙28 m

item of clothing

Calculate the centripetal force acting on the item of clothing

Space for working and answer

4

3

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2 (continued)

(d) The outer surface of the drum has small holes as shown in Figure 2C These holes allow most of the water to be removed

Figure 2C

(i) Explain why the water separates from the item of clothing during the spin cycle

(ii) The drum rotates in an anticlockwise direction

Indicate on Figure 2D the direction taken by a water droplet as it leaves the drum

Figure 2D

water droplet

drum

2

1

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2 (d) (continued)

(iii) Explain what happens to the magnitude of the force acting on an item of clothing inside the drum as it rotates at its maximum angular velocity

[Turn over

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3 A disc of mass 6∙0 kg and radius 0∙50 m is allowed to rotate freely about its central axis as shown in Figure 3A

central axis

Figure 3A

(a) Show that the moment of inertia of the disc is 0∙75 kg m2

Space for working and answer

(b) The disc is rotating with an angular velocity of 12 rad sminus1 A cube of mass 2∙0 kg is then dropped onto the disc The cube remains at a distance of 0∙40 m from the axis of rotation as shown in Figure 3B

central axis

040 m

Figure 3B

2

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3 (b) (continued)

(i) Determine the total moment of inertia of the disc and cube

Space for working and answer

(ii) Calculate the angular velocity of the disc after the cube lands

Space for working and answer

(iii) State one assumption you have made in your response to b(ii)

[Turn over

3

3

1

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3 (continued)

(c) The cube is removed and the disc is again made to rotate with a constant angular velocity of 12 rad sminus1 A sphere of mass 2∙0 kg is then dropped onto the disc at a distance of 0∙40 m from the axis as shown in Figure 3C

040 m

Figure 3C

State whether the resulting angular velocity of the disc is greater than the same as or less than the value calculated in b(ii)

Justify your answer 2

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4 The International Space Station (ISS) is in orbit around the Earth

ISS

40 times 105 m

not to scale

Figure 4A

(a) (i) The gravitational pull of the Earth keeps the ISS in orbit

Show that for an orbit of radius r the period T is given by the expression

3

2E

rT πGM

=

where the symbols have their usual meaning

(ii) Calculate the period of orbit of the ISS when it is at an altitude of 4∙0 times 105 m above the surface of the Earth

Space for working and answer

2

2

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4 (continued)

(b) The graph in Figure 4B shows how the altitude of the ISS has varied over time Reductions in altitude are due to the drag of the Earthrsquos atmosphere acting on the ISS

alti

tude

(times10

5 m

)

418

417

416

415

414

413

412

01 N

ov 2

013

01 D

ec 2

013

31 D

ec 2

013

30 J

an 2

014

01 M

ar 2

014

31 M

ar 2

014

30 A

pr 2

014

30 M

ay 2

014

29 J

un 2

014

29 J

ul 2

014

28 A

ug 2

014

27 S

ep 2

014

27 O

ct 2

014

26 N

ov 2

014

date

Figure 4B

(i) Determine the value of Earthrsquos gravitational field strength at the ISS on 1 March 2014

Space for working and answer

4

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4 (b) (continued)

(ii) In 2011 the average altitude of the ISS was increased from 350 km to 400 km

Suggest an advantage of operating the ISS at this higher altitude

(c) Clocks designed to operate on the ISS are synchronised with clocks on Earth before they go into space On the ISS a correction factor is necessary for the clocks to remain synchronised with clocks on Earth

Explain why this correction factor is necessary

[Turn over

1

2

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

S857770123page 23

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

S857770129page 29

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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THIS MARGIN

14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ 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2 (b) (continued)

(ii) Determine how many revolutions the drum will make during the 12∙0 seconds

Space for working and answer

(c) When the drum is rotating at maximum angular velocity an item of wet clothing of mass 1∙5 times 10minus2 kg rotates at a distance of 0∙28 m from the axis of rotation as shown in Figure 2B

Figure 2B

0∙28 m

item of clothing

Calculate the centripetal force acting on the item of clothing

Space for working and answer

4

3

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2 (continued)

(d) The outer surface of the drum has small holes as shown in Figure 2C These holes allow most of the water to be removed

Figure 2C

(i) Explain why the water separates from the item of clothing during the spin cycle

(ii) The drum rotates in an anticlockwise direction

Indicate on Figure 2D the direction taken by a water droplet as it leaves the drum

Figure 2D

water droplet

drum

2

1

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2 (d) (continued)

(iii) Explain what happens to the magnitude of the force acting on an item of clothing inside the drum as it rotates at its maximum angular velocity

[Turn over

2

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3 A disc of mass 6∙0 kg and radius 0∙50 m is allowed to rotate freely about its central axis as shown in Figure 3A

central axis

Figure 3A

(a) Show that the moment of inertia of the disc is 0∙75 kg m2

Space for working and answer

(b) The disc is rotating with an angular velocity of 12 rad sminus1 A cube of mass 2∙0 kg is then dropped onto the disc The cube remains at a distance of 0∙40 m from the axis of rotation as shown in Figure 3B

central axis

040 m

Figure 3B

2

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3 (b) (continued)

(i) Determine the total moment of inertia of the disc and cube

Space for working and answer

(ii) Calculate the angular velocity of the disc after the cube lands

Space for working and answer

(iii) State one assumption you have made in your response to b(ii)

[Turn over

3

3

1

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3 (continued)

(c) The cube is removed and the disc is again made to rotate with a constant angular velocity of 12 rad sminus1 A sphere of mass 2∙0 kg is then dropped onto the disc at a distance of 0∙40 m from the axis as shown in Figure 3C

040 m

Figure 3C

State whether the resulting angular velocity of the disc is greater than the same as or less than the value calculated in b(ii)

Justify your answer 2

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4 The International Space Station (ISS) is in orbit around the Earth

ISS

40 times 105 m

not to scale

Figure 4A

(a) (i) The gravitational pull of the Earth keeps the ISS in orbit

Show that for an orbit of radius r the period T is given by the expression

3

2E

rT πGM

=

where the symbols have their usual meaning

(ii) Calculate the period of orbit of the ISS when it is at an altitude of 4∙0 times 105 m above the surface of the Earth

Space for working and answer

2

2

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4 (continued)

(b) The graph in Figure 4B shows how the altitude of the ISS has varied over time Reductions in altitude are due to the drag of the Earthrsquos atmosphere acting on the ISS

alti

tude

(times10

5 m

)

418

417

416

415

414

413

412

01 N

ov 2

013

01 D

ec 2

013

31 D

ec 2

013

30 J

an 2

014

01 M

ar 2

014

31 M

ar 2

014

30 A

pr 2

014

30 M

ay 2

014

29 J

un 2

014

29 J

ul 2

014

28 A

ug 2

014

27 S

ep 2

014

27 O

ct 2

014

26 N

ov 2

014

date

Figure 4B

(i) Determine the value of Earthrsquos gravitational field strength at the ISS on 1 March 2014

Space for working and answer

4

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4 (b) (continued)

(ii) In 2011 the average altitude of the ISS was increased from 350 km to 400 km

Suggest an advantage of operating the ISS at this higher altitude

(c) Clocks designed to operate on the ISS are synchronised with clocks on Earth before they go into space On the ISS a correction factor is necessary for the clocks to remain synchronised with clocks on Earth

Explain why this correction factor is necessary

[Turn over

1

2

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

S857770130page 30

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THIS MARGIN

11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

S857770131page 31

MARKS DO NOT WRITE IN

THIS MARGIN

12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

S857770132page 32

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THIS MARGIN

12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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THIS MARGIN

13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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THIS MARGIN

14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

MARKS DO NOT WRITE IN

THIS MARGIN

14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

S857770139page 39

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ 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2 (continued)

(d) The outer surface of the drum has small holes as shown in Figure 2C These holes allow most of the water to be removed

Figure 2C

(i) Explain why the water separates from the item of clothing during the spin cycle

(ii) The drum rotates in an anticlockwise direction

Indicate on Figure 2D the direction taken by a water droplet as it leaves the drum

Figure 2D

water droplet

drum

2

1

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2 (d) (continued)

(iii) Explain what happens to the magnitude of the force acting on an item of clothing inside the drum as it rotates at its maximum angular velocity

[Turn over

2

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3 A disc of mass 6∙0 kg and radius 0∙50 m is allowed to rotate freely about its central axis as shown in Figure 3A

central axis

Figure 3A

(a) Show that the moment of inertia of the disc is 0∙75 kg m2

Space for working and answer

(b) The disc is rotating with an angular velocity of 12 rad sminus1 A cube of mass 2∙0 kg is then dropped onto the disc The cube remains at a distance of 0∙40 m from the axis of rotation as shown in Figure 3B

central axis

040 m

Figure 3B

2

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3 (b) (continued)

(i) Determine the total moment of inertia of the disc and cube

Space for working and answer

(ii) Calculate the angular velocity of the disc after the cube lands

Space for working and answer

(iii) State one assumption you have made in your response to b(ii)

[Turn over

3

3

1

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3 (continued)

(c) The cube is removed and the disc is again made to rotate with a constant angular velocity of 12 rad sminus1 A sphere of mass 2∙0 kg is then dropped onto the disc at a distance of 0∙40 m from the axis as shown in Figure 3C

040 m

Figure 3C

State whether the resulting angular velocity of the disc is greater than the same as or less than the value calculated in b(ii)

Justify your answer 2

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4 The International Space Station (ISS) is in orbit around the Earth

ISS

40 times 105 m

not to scale

Figure 4A

(a) (i) The gravitational pull of the Earth keeps the ISS in orbit

Show that for an orbit of radius r the period T is given by the expression

3

2E

rT πGM

=

where the symbols have their usual meaning

(ii) Calculate the period of orbit of the ISS when it is at an altitude of 4∙0 times 105 m above the surface of the Earth

Space for working and answer

2

2

[Turn over

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4 (continued)

(b) The graph in Figure 4B shows how the altitude of the ISS has varied over time Reductions in altitude are due to the drag of the Earthrsquos atmosphere acting on the ISS

alti

tude

(times10

5 m

)

418

417

416

415

414

413

412

01 N

ov 2

013

01 D

ec 2

013

31 D

ec 2

013

30 J

an 2

014

01 M

ar 2

014

31 M

ar 2

014

30 A

pr 2

014

30 M

ay 2

014

29 J

un 2

014

29 J

ul 2

014

28 A

ug 2

014

27 S

ep 2

014

27 O

ct 2

014

26 N

ov 2

014

date

Figure 4B

(i) Determine the value of Earthrsquos gravitational field strength at the ISS on 1 March 2014

Space for working and answer

4

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4 (b) (continued)

(ii) In 2011 the average altitude of the ISS was increased from 350 km to 400 km

Suggest an advantage of operating the ISS at this higher altitude

(c) Clocks designed to operate on the ISS are synchronised with clocks on Earth before they go into space On the ISS a correction factor is necessary for the clocks to remain synchronised with clocks on Earth

Explain why this correction factor is necessary

[Turn over

1

2

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

[Turn over

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP ltFEFF005500740069006c0069006300650020006500730074006100200063006f006e0066006900670075007200610063006900f3006e0020007000610072006100200063007200650061007200200064006f00630075006d0065006e0074006f00730020005000440046002000640065002000410064006f0062006500200061006400650063007500610064006f00730020007000610072006100200069006d0070007200650073006900f3006e0020007000720065002d0065006400690074006f007200690061006c00200064006500200061006c00740061002000630061006c0069006400610064002e002000530065002000700075006500640065006e00200061006200720069007200200064006f00630075006d0065006e0074006f00730020005000440046002000630072006500610064006f007300200063006f006e0020004100630072006f006200610074002c002000410064006f00620065002000520065006100640065007200200035002e003000200079002000760065007200730069006f006e0065007300200070006f00730074006500720069006f007200650073002egt FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName 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2 (d) (continued)

(iii) Explain what happens to the magnitude of the force acting on an item of clothing inside the drum as it rotates at its maximum angular velocity

[Turn over

2

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3 A disc of mass 6∙0 kg and radius 0∙50 m is allowed to rotate freely about its central axis as shown in Figure 3A

central axis

Figure 3A

(a) Show that the moment of inertia of the disc is 0∙75 kg m2

Space for working and answer

(b) The disc is rotating with an angular velocity of 12 rad sminus1 A cube of mass 2∙0 kg is then dropped onto the disc The cube remains at a distance of 0∙40 m from the axis of rotation as shown in Figure 3B

central axis

040 m

Figure 3B

2

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3 (b) (continued)

(i) Determine the total moment of inertia of the disc and cube

Space for working and answer

(ii) Calculate the angular velocity of the disc after the cube lands

Space for working and answer

(iii) State one assumption you have made in your response to b(ii)

[Turn over

3

3

1

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3 (continued)

(c) The cube is removed and the disc is again made to rotate with a constant angular velocity of 12 rad sminus1 A sphere of mass 2∙0 kg is then dropped onto the disc at a distance of 0∙40 m from the axis as shown in Figure 3C

040 m

Figure 3C

State whether the resulting angular velocity of the disc is greater than the same as or less than the value calculated in b(ii)

Justify your answer 2

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4 The International Space Station (ISS) is in orbit around the Earth

ISS

40 times 105 m

not to scale

Figure 4A

(a) (i) The gravitational pull of the Earth keeps the ISS in orbit

Show that for an orbit of radius r the period T is given by the expression

3

2E

rT πGM

=

where the symbols have their usual meaning

(ii) Calculate the period of orbit of the ISS when it is at an altitude of 4∙0 times 105 m above the surface of the Earth

Space for working and answer

2

2

[Turn over

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4 (continued)

(b) The graph in Figure 4B shows how the altitude of the ISS has varied over time Reductions in altitude are due to the drag of the Earthrsquos atmosphere acting on the ISS

alti

tude

(times10

5 m

)

418

417

416

415

414

413

412

01 N

ov 2

013

01 D

ec 2

013

31 D

ec 2

013

30 J

an 2

014

01 M

ar 2

014

31 M

ar 2

014

30 A

pr 2

014

30 M

ay 2

014

29 J

un 2

014

29 J

ul 2

014

28 A

ug 2

014

27 S

ep 2

014

27 O

ct 2

014

26 N

ov 2

014

date

Figure 4B

(i) Determine the value of Earthrsquos gravitational field strength at the ISS on 1 March 2014

Space for working and answer

4

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4 (b) (continued)

(ii) In 2011 the average altitude of the ISS was increased from 350 km to 400 km

Suggest an advantage of operating the ISS at this higher altitude

(c) Clocks designed to operate on the ISS are synchronised with clocks on Earth before they go into space On the ISS a correction factor is necessary for the clocks to remain synchronised with clocks on Earth

Explain why this correction factor is necessary

[Turn over

1

2

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

[Turn over

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

[Turn over

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

[Turn over

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF00560065007200770065006e00640065006e0020005300690065002000640069006500730065002000450069006e007300740065006c006c0075006e00670065006e0020007a0075006d002000450072007300740065006c006c0065006e00200076006f006e002000410064006f006200650020005000440046002d0044006f006b0075006d0065006e00740065006e002c00200076006f006e002000640065006e0065006e002000530069006500200068006f006300680077006500720074006900670065002000500072006500700072006500730073002d0044007200750063006b0065002000650072007a0065007500670065006e0020006d00f60063006800740065006e002e002000450072007300740065006c006c007400650020005000440046002d0044006f006b0075006d0065006e007400650020006b00f6006e006e0065006e0020006d006900740020004100630072006f00620061007400200075006e0064002000410064006f00620065002000520065006100640065007200200035002e00300020006f0064006500720020006800f600680065007200200067006500f600660066006e00650074002000770065007200640065006e002egt ESP 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 FRA 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 ITA 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 JPN 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S857770108page 08

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3 A disc of mass 6∙0 kg and radius 0∙50 m is allowed to rotate freely about its central axis as shown in Figure 3A

central axis

Figure 3A

(a) Show that the moment of inertia of the disc is 0∙75 kg m2

Space for working and answer

(b) The disc is rotating with an angular velocity of 12 rad sminus1 A cube of mass 2∙0 kg is then dropped onto the disc The cube remains at a distance of 0∙40 m from the axis of rotation as shown in Figure 3B

central axis

040 m

Figure 3B

2

S857770109page 09

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3 (b) (continued)

(i) Determine the total moment of inertia of the disc and cube

Space for working and answer

(ii) Calculate the angular velocity of the disc after the cube lands

Space for working and answer

(iii) State one assumption you have made in your response to b(ii)

[Turn over

3

3

1

S857770110page 10

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3 (continued)

(c) The cube is removed and the disc is again made to rotate with a constant angular velocity of 12 rad sminus1 A sphere of mass 2∙0 kg is then dropped onto the disc at a distance of 0∙40 m from the axis as shown in Figure 3C

040 m

Figure 3C

State whether the resulting angular velocity of the disc is greater than the same as or less than the value calculated in b(ii)

Justify your answer 2

S857770111page 11

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4 The International Space Station (ISS) is in orbit around the Earth

ISS

40 times 105 m

not to scale

Figure 4A

(a) (i) The gravitational pull of the Earth keeps the ISS in orbit

Show that for an orbit of radius r the period T is given by the expression

3

2E

rT πGM

=

where the symbols have their usual meaning

(ii) Calculate the period of orbit of the ISS when it is at an altitude of 4∙0 times 105 m above the surface of the Earth

Space for working and answer

2

2

[Turn over

S857770112page 12

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4 (continued)

(b) The graph in Figure 4B shows how the altitude of the ISS has varied over time Reductions in altitude are due to the drag of the Earthrsquos atmosphere acting on the ISS

alti

tude

(times10

5 m

)

418

417

416

415

414

413

412

01 N

ov 2

013

01 D

ec 2

013

31 D

ec 2

013

30 J

an 2

014

01 M

ar 2

014

31 M

ar 2

014

30 A

pr 2

014

30 M

ay 2

014

29 J

un 2

014

29 J

ul 2

014

28 A

ug 2

014

27 S

ep 2

014

27 O

ct 2

014

26 N

ov 2

014

date

Figure 4B

(i) Determine the value of Earthrsquos gravitational field strength at the ISS on 1 March 2014

Space for working and answer

4

S857770113page 13

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4 (b) (continued)

(ii) In 2011 the average altitude of the ISS was increased from 350 km to 400 km

Suggest an advantage of operating the ISS at this higher altitude

(c) Clocks designed to operate on the ISS are synchronised with clocks on Earth before they go into space On the ISS a correction factor is necessary for the clocks to remain synchronised with clocks on Earth

Explain why this correction factor is necessary

[Turn over

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2

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

[Turn over

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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3 (b) (continued)

(i) Determine the total moment of inertia of the disc and cube

Space for working and answer

(ii) Calculate the angular velocity of the disc after the cube lands

Space for working and answer

(iii) State one assumption you have made in your response to b(ii)

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1

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3 (continued)

(c) The cube is removed and the disc is again made to rotate with a constant angular velocity of 12 rad sminus1 A sphere of mass 2∙0 kg is then dropped onto the disc at a distance of 0∙40 m from the axis as shown in Figure 3C

040 m

Figure 3C

State whether the resulting angular velocity of the disc is greater than the same as or less than the value calculated in b(ii)

Justify your answer 2

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4 The International Space Station (ISS) is in orbit around the Earth

ISS

40 times 105 m

not to scale

Figure 4A

(a) (i) The gravitational pull of the Earth keeps the ISS in orbit

Show that for an orbit of radius r the period T is given by the expression

3

2E

rT πGM

=

where the symbols have their usual meaning

(ii) Calculate the period of orbit of the ISS when it is at an altitude of 4∙0 times 105 m above the surface of the Earth

Space for working and answer

2

2

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4 (continued)

(b) The graph in Figure 4B shows how the altitude of the ISS has varied over time Reductions in altitude are due to the drag of the Earthrsquos atmosphere acting on the ISS

alti

tude

(times10

5 m

)

418

417

416

415

414

413

412

01 N

ov 2

013

01 D

ec 2

013

31 D

ec 2

013

30 J

an 2

014

01 M

ar 2

014

31 M

ar 2

014

30 A

pr 2

014

30 M

ay 2

014

29 J

un 2

014

29 J

ul 2

014

28 A

ug 2

014

27 S

ep 2

014

27 O

ct 2

014

26 N

ov 2

014

date

Figure 4B

(i) Determine the value of Earthrsquos gravitational field strength at the ISS on 1 March 2014

Space for working and answer

4

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4 (b) (continued)

(ii) In 2011 the average altitude of the ISS was increased from 350 km to 400 km

Suggest an advantage of operating the ISS at this higher altitude

(c) Clocks designed to operate on the ISS are synchronised with clocks on Earth before they go into space On the ISS a correction factor is necessary for the clocks to remain synchronised with clocks on Earth

Explain why this correction factor is necessary

[Turn over

1

2

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

S857770127page 27

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THIS MARGIN

9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

S857770129page 29

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THIS MARGIN

10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

S857770131page 31

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName 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3 (continued)

(c) The cube is removed and the disc is again made to rotate with a constant angular velocity of 12 rad sminus1 A sphere of mass 2∙0 kg is then dropped onto the disc at a distance of 0∙40 m from the axis as shown in Figure 3C

040 m

Figure 3C

State whether the resulting angular velocity of the disc is greater than the same as or less than the value calculated in b(ii)

Justify your answer 2

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4 The International Space Station (ISS) is in orbit around the Earth

ISS

40 times 105 m

not to scale

Figure 4A

(a) (i) The gravitational pull of the Earth keeps the ISS in orbit

Show that for an orbit of radius r the period T is given by the expression

3

2E

rT πGM

=

where the symbols have their usual meaning

(ii) Calculate the period of orbit of the ISS when it is at an altitude of 4∙0 times 105 m above the surface of the Earth

Space for working and answer

2

2

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4 (continued)

(b) The graph in Figure 4B shows how the altitude of the ISS has varied over time Reductions in altitude are due to the drag of the Earthrsquos atmosphere acting on the ISS

alti

tude

(times10

5 m

)

418

417

416

415

414

413

412

01 N

ov 2

013

01 D

ec 2

013

31 D

ec 2

013

30 J

an 2

014

01 M

ar 2

014

31 M

ar 2

014

30 A

pr 2

014

30 M

ay 2

014

29 J

un 2

014

29 J

ul 2

014

28 A

ug 2

014

27 S

ep 2

014

27 O

ct 2

014

26 N

ov 2

014

date

Figure 4B

(i) Determine the value of Earthrsquos gravitational field strength at the ISS on 1 March 2014

Space for working and answer

4

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4 (b) (continued)

(ii) In 2011 the average altitude of the ISS was increased from 350 km to 400 km

Suggest an advantage of operating the ISS at this higher altitude

(c) Clocks designed to operate on the ISS are synchronised with clocks on Earth before they go into space On the ISS a correction factor is necessary for the clocks to remain synchronised with clocks on Earth

Explain why this correction factor is necessary

[Turn over

1

2

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

[Turn over

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

S857770111page 11

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4 The International Space Station (ISS) is in orbit around the Earth

ISS

40 times 105 m

not to scale

Figure 4A

(a) (i) The gravitational pull of the Earth keeps the ISS in orbit

Show that for an orbit of radius r the period T is given by the expression

3

2E

rT πGM

=

where the symbols have their usual meaning

(ii) Calculate the period of orbit of the ISS when it is at an altitude of 4∙0 times 105 m above the surface of the Earth

Space for working and answer

2

2

[Turn over

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4 (continued)

(b) The graph in Figure 4B shows how the altitude of the ISS has varied over time Reductions in altitude are due to the drag of the Earthrsquos atmosphere acting on the ISS

alti

tude

(times10

5 m

)

418

417

416

415

414

413

412

01 N

ov 2

013

01 D

ec 2

013

31 D

ec 2

013

30 J

an 2

014

01 M

ar 2

014

31 M

ar 2

014

30 A

pr 2

014

30 M

ay 2

014

29 J

un 2

014

29 J

ul 2

014

28 A

ug 2

014

27 S

ep 2

014

27 O

ct 2

014

26 N

ov 2

014

date

Figure 4B

(i) Determine the value of Earthrsquos gravitational field strength at the ISS on 1 March 2014

Space for working and answer

4

S857770113page 13

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4 (b) (continued)

(ii) In 2011 the average altitude of the ISS was increased from 350 km to 400 km

Suggest an advantage of operating the ISS at this higher altitude

(c) Clocks designed to operate on the ISS are synchronised with clocks on Earth before they go into space On the ISS a correction factor is necessary for the clocks to remain synchronised with clocks on Earth

Explain why this correction factor is necessary

[Turn over

1

2

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

S857770115page 15

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

[Turn over

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

S857770119page 19

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

S857770125page 25

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

S857770127page 27

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

[Turn over

S857770128page 28

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

S857770129page 29

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

S857770130page 30

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

S857770131page 31

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

S857770132page 32

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

S857770134page 34

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

S857770139page 39

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 ESP 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 SVE 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PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

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4 (continued)

(b) The graph in Figure 4B shows how the altitude of the ISS has varied over time Reductions in altitude are due to the drag of the Earthrsquos atmosphere acting on the ISS

alti

tude

(times10

5 m

)

418

417

416

415

414

413

412

01 N

ov 2

013

01 D

ec 2

013

31 D

ec 2

013

30 J

an 2

014

01 M

ar 2

014

31 M

ar 2

014

30 A

pr 2

014

30 M

ay 2

014

29 J

un 2

014

29 J

ul 2

014

28 A

ug 2

014

27 S

ep 2

014

27 O

ct 2

014

26 N

ov 2

014

date

Figure 4B

(i) Determine the value of Earthrsquos gravitational field strength at the ISS on 1 March 2014

Space for working and answer

4

S857770113page 13

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4 (b) (continued)

(ii) In 2011 the average altitude of the ISS was increased from 350 km to 400 km

Suggest an advantage of operating the ISS at this higher altitude

(c) Clocks designed to operate on the ISS are synchronised with clocks on Earth before they go into space On the ISS a correction factor is necessary for the clocks to remain synchronised with clocks on Earth

Explain why this correction factor is necessary

[Turn over

1

2

S857770114page 14

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

S857770115page 15

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

[Turn over

S857770116page 16

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

S857770117page 17

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

[Turn over

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

S857770119page 19

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

[Turn over

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

[Turn over

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

S857770123page 23

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

S857770124page 24

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THIS MARGIN

8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

S857770125page 25

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

[Turn over

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

S857770127page 27

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THIS MARGIN

9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

[Turn over

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

S857770129page 29

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THIS MARGIN

10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

S857770134page 34

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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THIS MARGIN

14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

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4 (b) (continued)

(ii) In 2011 the average altitude of the ISS was increased from 350 km to 400 km

Suggest an advantage of operating the ISS at this higher altitude

(c) Clocks designed to operate on the ISS are synchronised with clocks on Earth before they go into space On the ISS a correction factor is necessary for the clocks to remain synchronised with clocks on Earth

Explain why this correction factor is necessary

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

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1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

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L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA ltFEFF005500740069006c006900730065007a00200063006500730020006f007000740069006f006e00730020006100660069006e00200064006500200063007200e900650072002000640065007300200064006f00630075006d0065006e00740073002000410064006f00620065002000500044004600200070006f0075007200200075006e00650020007100750061006c0069007400e90020006400270069006d007000720065007300730069006f006e00200070007200e9007000720065007300730065002e0020004c0065007300200064006f00630075006d0065006e00740073002000500044004600200063007200e900e90073002000700065007500760065006e0074002000ea0074007200650020006f007500760065007200740073002000640061006e00730020004100630072006f006200610074002c002000610069006e00730069002000710075002700410064006f00620065002000520065006100640065007200200035002e0030002000650074002000760065007200730069006f006e007300200075006c007400e90072006900650075007200650073002egt ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

S857770114page 14

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5 The constellation Orion shown in Figure 5A is a common sight in the winter sky above Scotland

BetelgeuseMeissa

Bellatrix

Mintaka

Alnitak Alnilam

Rigel

Saiph

Figure 5A

Two of the stars in this constellation are known as Betelgeuse and Rigel Their positions are shown on the HertzsprungndashRussell (H-R) diagram in Figure 5B

S857770115page 15

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

[Turn over

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

S857770134page 34

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

S857770139page 39

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ 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S857770115page 15

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5 (continued)

Rigel

RegulusVega

Deneb

Betelgeuse

Antares

ArcturusAldebaran

Mira

PolluxProcyon A

Procyon B

BarnardrsquosStar

Sun

Sirius A

Sirius B

Altair

25 000 10 000 8000 6000 5000 4000 3000

106 minus10

minus5

0

+15

+10

104

102

1

abso

lute

mag

nitu

de10ndash2

10ndash4

lum

inos

ity

(Sun

= 1

)surface temperature (K)

spectral type

O5 B0 A0 F0 G0 K0 M0 M8

+5

Figure 5B

(a) Using the H-R diagram predict the colour of Betelgeuse 1

[Turn over

S857770116page 16

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

S857770117page 17

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

[Turn over

S857770118page 18

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

S857770119page 19

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

[Turn over

S857770120page 20

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

S857770121page 21

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

[Turn over

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

S857770123page 23

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

S857770124page 24

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

[Turn over

S857770126page 26

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

S857770127page 27

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

S857770129page 29

MARKS DO NOT WRITE IN

THIS MARGIN

10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

S857770131page 31

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THIS MARGIN

12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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THIS MARGIN

13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

MARKS DO NOT WRITE IN

THIS MARGIN

14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE ltFEFF0041006e007600e4006e00640020006400650020006800e4007200200069006e0073007400e4006c006c006e0069006e006700610072006e00610020006f006d002000640075002000760069006c006c00200073006b006100700061002000410064006f006200650020005000440046002d0064006f006b0075006d0065006e007400200073006f006d002000e400720020006c00e4006d0070006c0069006700610020006600f60072002000700072006500700072006500730073002d007500740073006b00720069006600740020006d006500640020006800f600670020006b00760061006c0069007400650074002e002000200053006b006100700061006400650020005000440046002d0064006f006b0075006d0065006e00740020006b0061006e002000f600700070006e00610073002000690020004100630072006f0062006100740020006f00630068002000410064006f00620065002000520065006100640065007200200035002e00300020006f00630068002000730065006e006100720065002egt ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

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5 (continued)

(b) The table shows some of the physical properties of Rigel

Property of Rigel

Surface temperature (1∙20 plusmn 0∙05) times 104 K

Radius (5∙49 plusmn 0∙50) times 1010 m

Mass 18 plusmn 1 solar masses

Distance to Earth 773 plusmn 150 light years

(i) (A) Calculate the luminosity of Rigel

Space for working and answer

(B) State the assumption made in your calculation

3

1

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 ESP 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 ITA 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 JPN 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S857770117page 17

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5 (b) (continued)

(ii) Calculate the absolute uncertainty in the value of the luminosity of Rigel

Space for working and answer

(c) Calculate the apparent brightness of Rigel as observed from the Earth

Space for working and answer

(d) Betelgeuse is not on the Main Sequence region of the H-R diagram

State two changes that took place in the core of Betelgeuse at the point it left the Main Sequence

(e) Explain in terms of thermal pressure and gravitational force the change in the radius of Betelgeuse between leaving the Main Sequence and becoming a supergiant

4

4

2

2

[Turn over

S857770118page 18

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

S857770119page 19

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THIS MARGIN

6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

[Turn over

S857770120page 20

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

S857770121page 21

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

[Turn over

S857770122page 22

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

S857770123page 23

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

S857770125page 25

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THIS MARGIN

8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

S857770127page 27

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THIS MARGIN

9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

S857770129page 29

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THIS MARGIN

10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

S857770131page 31

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THIS MARGIN

12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP ltFEFF005500740069006c0069006300650020006500730074006100200063006f006e0066006900670075007200610063006900f3006e0020007000610072006100200063007200650061007200200064006f00630075006d0065006e0074006f00730020005000440046002000640065002000410064006f0062006500200061006400650063007500610064006f00730020007000610072006100200069006d0070007200650073006900f3006e0020007000720065002d0065006400690074006f007200690061006c00200064006500200061006c00740061002000630061006c0069006400610064002e002000530065002000700075006500640065006e00200061006200720069007200200064006f00630075006d0065006e0074006f00730020005000440046002000630072006500610064006f007300200063006f006e0020004100630072006f006200610074002c002000410064006f00620065002000520065006100640065007200200035002e003000200079002000760065007200730069006f006e0065007300200070006f00730074006500720069006f007200650073002egt FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

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6 The Bohr model of the hydrogen atom consists of a single electron orbiting a single proton Due to the quantisation of angular momentum in this model the electron can only orbit at particular radii

Figure 6A shows an electron with principal quantum number n = 1

n = 2

n = 1electron

Figure 6A

(a) Explain what gives rise to the centripetal force acting on the electron

(b) (i) Show that the kinetic energy of the electron is given by

π

2

08keEε r

=

where the symbols have their usual meaning

1

2

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 ESP 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 JPN 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S857770119page 19

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6 (b) (continued)

(ii) Calculate the kinetic energy for an electron with orbital radius 0∙21 nm

Space for working and answer

(c) Calculate the principal quantum number for an electron with angular momentum 4∙22 times 10minus34 kg m2 sminus1

Space for working and answer

(d) The Heisenberg uncertainty principle addresses some of the limitations of classical physics in describing quantum phenomena

(i) The uncertainty in an experimental measurement of the momentum of an electron in a hydrogen atom was determined to be plusmn1∙5 times 10minus26 kg m sminus1

Calculate the minimum uncertainty Δxmin in the position of the electron

Space for working and answer

2

3

3

[Turn over

S857770120page 20

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

S857770121page 21

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

[Turn over

S857770122page 22

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

S857770123page 23

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THIS MARGIN

8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

S857770125page 25

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THIS MARGIN

8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

S857770127page 27

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THIS MARGIN

9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

S857770129page 29

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THIS MARGIN

10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

S857770131page 31

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

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6 (d) (continued)

(ii) In a scanning tunnelling microscope (STM) a sharp metallic tip is brought very close to the surface of a conductor As the tip is moved back and forth an electric current can be detected due to the movement (lsquotunnellingrsquo) of electrons across the gap between the tip and the conductor as shown in Figure 6B

STM tip

tunnelling electrons

Figure 6B

According to classical physics electrons should not be able to cross the gap as the kinetic energy of each electron is insufficient to overcome the repulsion between electrons in the STM tip and the surface

Explain why an electron is able to cross the gap 3

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 ESP 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 ITA 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 JPN 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S857770121page 21

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7 A proton travelling with a speed of 2middot6 times 107 m sminus1 enters the magnetic field around the Earth at an angle of 50deg as shown in Figure 7B The magnetic field strength is 58 μT

protonpath

50deg

B

Figure 7B

(a) Explain the shape of the path followed by the proton in the magnetic field

(b) Calculate the radius of curvature of this path

Space for working and answer

2

5

[Turn over

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

S857770123page 23

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

S857770124page 24

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

S857770125page 25

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

[Turn over

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

S857770127page 27

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

[Turn over

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

S857770129page 29

MARKS DO NOT WRITE IN

THIS MARGIN

10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

S857770130page 30

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THIS MARGIN

11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

S857770131page 31

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THIS MARGIN

12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

S857770132page 32

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THIS MARGIN

12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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THIS MARGIN

13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

S857770134page 34

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THIS MARGIN

14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

MARKS DO NOT WRITE IN

THIS MARGIN

14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

S857770139page 39

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO ltFEFF004b00e40079007400e40020006e00e40069007400e4002000610073006500740075006b007300690061002c0020006b0075006e0020006c0075006f00740020006c00e400680069006e006e00e4002000760061006100740069007600610061006e0020007000610069006e006100740075006b00730065006e002000760061006c006d0069007300740065006c00750074007900f6006800f6006e00200073006f00700069007600690061002000410064006f0062006500200050004400460020002d0064006f006b0075006d0065006e007400740065006a0061002e0020004c0075006f0064007500740020005000440046002d0064006f006b0075006d0065006e00740069007400200076006f0069006400610061006e0020006100760061007400610020004100630072006f0062006100740069006c006c00610020006a0061002000410064006f00620065002000520065006100640065007200200035002e0030003a006c006c00610020006a006100200075007500640065006d006d0069006c006c0061002egt SVE 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(Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

S857770122page 22

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7 (continued)

(c) An antiproton travelling with a speed of 2middot6 times 107 m sminus1 enters the same region of the Earthrsquos magnetic field at an angle of 30deg to the field

Describe two differences in the paths taken by the antiproton and the original proton 2

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

S857770129page 29

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

S857770130page 30

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

S857770134page 34

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

S857770139page 39

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn 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 SUO ltFEFF004b00e40079007400e40020006e00e40069007400e4002000610073006500740075006b007300690061002c0020006b0075006e0020006c0075006f00740020006c00e400680069006e006e00e4002000760061006100740069007600610061006e0020007000610069006e006100740075006b00730065006e002000760061006c006d0069007300740065006c00750074007900f6006800f6006e00200073006f00700069007600690061002000410064006f0062006500200050004400460020002d0064006f006b0075006d0065006e007400740065006a0061002e0020004c0075006f0064007500740020005000440046002d0064006f006b0075006d0065006e00740069007400200076006f0069006400610061006e0020006100760061007400610020004100630072006f0062006100740069006c006c00610020006a0061002000410064006f00620065002000520065006100640065007200200035002e0030003a006c006c00610020006a006100200075007500640065006d006d0069006c006c0061002egt SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

S857770123page 23

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8 Figure 8A shows a snowboarder in a half pipe The snowboarder is moving between positions P and Q The total mass of snowboarder and board is 85 kg

P Q

OFigure 8A

A student attempts to model the motion of the snowboarder in the horizontal plane as simple harmonic motion (SHM)

The student uses measurements of amplitude and period to produce the displacement-time graph shown in Figure 8B

disp

lace

men

t (m

)

time (s)10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

50

40

30

20

10

minus10

minus20

minus30

minus40

minus50

0

Figure 8B

(a) (i) State what is meant by the term simple harmonic motion

[Turn over

1

S857770124page 24

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

S857770125page 25

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

[Turn over

S857770126page 26

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

S857770127page 27

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

[Turn over

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

S857770129page 29

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

S857770130page 30

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

S857770132page 32

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

S857770134page 34

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

S857770139page 39

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN 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geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

S857770124page 24

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8 (a) (continued)

(ii) Determine the angular frequency of the motion

Space for working and answer

(iii) Calculate the maximum acceleration in the horizontal plane experienced by the snowboarder

Space for working and answer

(iv) Sketch a velocity-time graph for one period of the motion in the horizontal plane

Numerical values are required on both axes

You may wish to use the square-ruled paper on page 36

4

3

3

S857770125page 25

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

S857770129page 29

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

S857770130page 30

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

S857770132page 32

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

S857770134page 34

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

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8 (a) (continued)

(v) Calculate the maximum potential energy of the snowboarder

Space for working and answer

(b) Detailed video analysis shows that the snowboarderrsquos motion is not fully described by the SHM model

Using your knowledge of physics comment on possible reasons for this discrepancy

3

3

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9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn 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 SUO 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 SVE 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S857770126page 26

MARKS DO NOT WRITE IN

THIS MARGIN

9 When a microwave oven is switched on a stationary wave is formed inside the oven

(a) Explain how a stationary wave is formed

(b) A student carries out an experiment to determine the speed of light using a microwave oven The turntable is removed from the oven and bread covered in butter is placed inside The oven is switched on for a short time after which the student observes that the butter has melted only in certain spots as shown in Figure 9A

melted spot

Figure 9A

Explain why the butter has melted in certain spots and not in others

1

2

S857770127page 27

MARKS DO NOT WRITE IN

THIS MARGIN

9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

S857770129page 29

MARKS DO NOT WRITE IN

THIS MARGIN

10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

S857770131page 31

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THIS MARGIN

12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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THIS MARGIN

13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

MARKS DO NOT WRITE IN

THIS MARGIN

14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU ltFEFF00560065007200770065006e00640065006e0020005300690065002000640069006500730065002000450069006e007300740065006c006c0075006e00670065006e0020007a0075006d002000450072007300740065006c006c0065006e00200076006f006e002000410064006f006200650020005000440046002d0044006f006b0075006d0065006e00740065006e002c00200076006f006e002000640065006e0065006e002000530069006500200068006f006300680077006500720074006900670065002000500072006500700072006500730073002d0044007200750063006b0065002000650072007a0065007500670065006e0020006d00f60063006800740065006e002e002000450072007300740065006c006c007400650020005000440046002d0044006f006b0075006d0065006e007400650020006b00f6006e006e0065006e0020006d006900740020004100630072006f00620061007400200075006e0064002000410064006f00620065002000520065006100640065007200200035002e00300020006f0064006500720020006800f600680065007200200067006500f600660066006e00650074002000770065007200640065006e002egt ESP ltFEFF005500740069006c0069006300650020006500730074006100200063006f006e0066006900670075007200610063006900f3006e0020007000610072006100200063007200650061007200200064006f00630075006d0065006e0074006f00730020005000440046002000640065002000410064006f0062006500200061006400650063007500610064006f00730020007000610072006100200069006d0070007200650073006900f3006e0020007000720065002d0065006400690074006f007200690061006c00200064006500200061006c00740061002000630061006c0069006400610064002e002000530065002000700075006500640065006e00200061006200720069007200200064006f00630075006d0065006e0074006f00730020005000440046002000630072006500610064006f007300200063006f006e0020004100630072006f006200610074002c002000410064006f00620065002000520065006100640065007200200035002e003000200079002000760065007200730069006f006e0065007300200070006f00730074006500720069006f007200650073002egt FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

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9 (continued)

(c) The student measures the distance between the first melted spot and fifth melted spot as 264 mm

264 mm

Figure 9B

From the data obtained by the student determine the wavelength of the microwaves

Space for working and answer

(d)  The quoted value for the frequency of the microwaves is 2∙45 GHz The student calculates the speed of light using data from the experiment

Show that the value obtained by the student for the speed of light is 3∙23 times 108 m sminus1

Space for working and answer

2

2

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 ITA 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 JPN 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 SVE 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PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

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9 (continued)

(e) The student repeats the experiment and obtains the following values for the speed of light

3∙26 times 108 m sminus1 3∙19 times 108 m sminus1 3∙23 times 108 m sminus1 3∙21 times 108 m sminus1

Comment on both the accuracy and precision of the studentrsquos results 2

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10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

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11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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S857770129page 29

MARKS DO NOT WRITE IN

THIS MARGIN

10 A beam of electrons is incident on a grating as shown in Figure 10A

grating

electron gun

zinc sulfide coated screen

Figure 10A

(a) After passing through the grating the electrons are incident on a zinc sulfide coated screen The coating emits light when struck by electrons

Describe the pattern observed on the screen

(b) Scientists perform similar experiments with large molecules One such molecule is buckminsterfullerene (C60) with a mass of 1∙20 times 10minus24 kg

For C60 molecules with a velocity of 220 m sminus1 estimate the slit spacing required to produce a pattern comparable to that observed for the electrons

You must justify your answer by calculation

Space for working and answer

1

4

[Turn over

S857770130page 30

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THIS MARGIN

11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

S857770131page 31

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THIS MARGIN

12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

S857770132page 32

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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THIS MARGIN

14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

S857770130page 30

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THIS MARGIN

11 As part of a physics project a student carried out experiments to obtain values for the permeability of free space and the permittivity of free space

The results obtained by the student were

permeability of free space ( ) 6 10 1middot32 0middot05 10 Hmμ minus minus= plusmn times

permittivity of free space ( ) 12 10 8middot93 0middot07 10 Fmε minus minus= plusmn times

(a) State the number of significant figures in the value of each result

(b) Use these results to determine a value for the speed of light

Space for working and answer

(c) (i) Determine which of the uncertainties obtained by the student is more significant for the calculation of the speed of light

You must justify your answer by calculation

Space for working and answer

(ii) Calculate the absolute uncertainty in the value obtained for the speed of light

Space for working and answer

1

3

3

2

S857770131page 31

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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

[Turn over

S857770132page 32

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

S857770134page 34

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

S857770139page 39

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 ESP 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12 (a) Two point charges Q1 and Q2 are separated by a distance of 0∙60 m as shown in Figure 12A

The charge on Q1 is minus8∙0 nC The electric field strength at point X is zero 

040 m

minus80 nC XQ2Q1

020 m

Figure 12A

(i) State what is meant by electric field strength

(ii) Show that the charge on Q2 is minus2∙0 nC

Space for working and answer

   (iii)  Determine the electrical potential at point X

Space for working and answer

1

2

5

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12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU ltFEFF00560065007200770065006e00640065006e0020005300690065002000640069006500730065002000450069006e007300740065006c006c0075006e00670065006e0020007a0075006d002000450072007300740065006c006c0065006e00200076006f006e002000410064006f006200650020005000440046002d0044006f006b0075006d0065006e00740065006e002c00200076006f006e002000640065006e0065006e002000530069006500200068006f006300680077006500720074006900670065002000500072006500700072006500730073002d0044007200750063006b0065002000650072007a0065007500670065006e0020006d00f60063006800740065006e002e002000450072007300740065006c006c007400650020005000440046002d0044006f006b0075006d0065006e007400650020006b00f6006e006e0065006e0020006d006900740020004100630072006f00620061007400200075006e0064002000410064006f00620065002000520065006100640065007200200035002e00300020006f0064006500720020006800f600680065007200200067006500f600660066006e00650074002000770065007200640065006e002egt ESP ltFEFF005500740069006c0069006300650020006500730074006100200063006f006e0066006900670075007200610063006900f3006e0020007000610072006100200063007200650061007200200064006f00630075006d0065006e0074006f00730020005000440046002000640065002000410064006f0062006500200061006400650063007500610064006f00730020007000610072006100200069006d0070007200650073006900f3006e0020007000720065002d0065006400690074006f007200690061006c00200064006500200061006c00740061002000630061006c0069006400610064002e002000530065002000700075006500640065006e00200061006200720069007200200064006f00630075006d0065006e0074006f00730020005000440046002000630072006500610064006f007300200063006f006e0020004100630072006f006200610074002c002000410064006f00620065002000520065006100640065007200200035002e003000200079002000760065007200730069006f006e0065007300200070006f00730074006500720069006f007200650073002egt FRA ltFEFF005500740069006c006900730065007a00200063006500730020006f007000740069006f006e00730020006100660069006e00200064006500200063007200e900650072002000640065007300200064006f00630075006d0065006e00740073002000410064006f00620065002000500044004600200070006f0075007200200075006e00650020007100750061006c0069007400e90020006400270069006d007000720065007300730069006f006e00200070007200e9007000720065007300730065002e0020004c0065007300200064006f00630075006d0065006e00740073002000500044004600200063007200e900e90073002000700065007500760065006e0074002000ea0074007200650020006f007500760065007200740073002000640061006e00730020004100630072006f006200610074002c002000610069006e00730069002000710075002700410064006f00620065002000520065006100640065007200200035002e0030002000650074002000760065007200730069006f006e007300200075006c007400e90072006900650075007200650073002egt ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR ltFEFF004200720075006b00200064006900730073006500200069006e006e007300740069006c006c0069006e00670065006e0065002000740069006c002000e50020006f0070007000720065007400740065002000410064006f006200650020005000440046002d0064006f006b0075006d0065006e00740065007200200073006f006d00200065007200200062006500730074002000650067006e0065007400200066006f00720020006600f80072007400720079006b006b0073007500740073006b00720069006600740020006100760020006800f800790020006b00760061006c0069007400650074002e0020005000440046002d0064006f006b0075006d0065006e00740065006e00650020006b0061006e002000e50070006e00650073002000690020004100630072006f00620061007400200065006c006c00650072002000410064006f00620065002000520065006100640065007200200035002e003000200065006c006c00650072002000730065006e006500720065002egt PTB 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 SUO ltFEFF004b00e40079007400e40020006e00e40069007400e4002000610073006500740075006b007300690061002c0020006b0075006e0020006c0075006f00740020006c00e400680069006e006e00e4002000760061006100740069007600610061006e0020007000610069006e006100740075006b00730065006e002000760061006c006d0069007300740065006c00750074007900f6006800f6006e00200073006f00700069007600690061002000410064006f0062006500200050004400460020002d0064006f006b0075006d0065006e007400740065006a0061002e0020004c0075006f0064007500740020005000440046002d0064006f006b0075006d0065006e00740069007400200076006f0069006400610061006e0020006100760061007400610020004100630072006f0062006100740069006c006c00610020006a0061002000410064006f00620065002000520065006100640065007200200035002e0030003a006c006c00610020006a006100200075007500640065006d006d0069006c006c0061002egt SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

S857770132page 32

MARKS DO NOT WRITE IN

THIS MARGIN

12 (continued)

(b)

040 m 040 m

minus20 nCminus80 nC

P

X

030 m030 m

Q2Q1

040 m 020 m

Figure 12B

(i) Determine the electrical potential at point P

Space for working and answer

(ii) Determine the energy required to move a charge of +1∙0 nC from point X to point P

Space for working and answer

3

4

S857770133page 33

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13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

S857770134page 34

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14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

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14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

S857770139page 39

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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ltFEFF00560065007200770065006e00640065006e0020005300690065002000640069006500730065002000450069006e007300740065006c006c0075006e00670065006e0020007a0075006d002000450072007300740065006c006c0065006e00200076006f006e002000410064006f006200650020005000440046002d0044006f006b0075006d0065006e00740065006e002c00200076006f006e002000640065006e0065006e002000530069006500200068006f006300680077006500720074006900670065002000500072006500700072006500730073002d0044007200750063006b0065002000650072007a0065007500670065006e0020006d00f60063006800740065006e002e002000450072007300740065006c006c007400650020005000440046002d0044006f006b0075006d0065006e007400650020006b00f6006e006e0065006e0020006d006900740020004100630072006f00620061007400200075006e0064002000410064006f00620065002000520065006100640065007200200035002e00300020006f0064006500720020006800f600680065007200200067006500f600660066006e00650074002000770065007200640065006e002egt ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR ltFEFF004200720075006b00200064006900730073006500200069006e006e007300740069006c006c0069006e00670065006e0065002000740069006c002000e50020006f0070007000720065007400740065002000410064006f006200650020005000440046002d0064006f006b0075006d0065006e00740065007200200073006f006d00200065007200200062006500730074002000650067006e0065007400200066006f00720020006600f80072007400720079006b006b0073007500740073006b00720069006600740020006100760020006800f800790020006b00760061006c0069007400650074002e0020005000440046002d0064006f006b0075006d0065006e00740065006e00650020006b0061006e002000e50070006e00650073002000690020004100630072006f00620061007400200065006c006c00650072002000410064006f00620065002000520065006100640065007200200035002e003000200065006c006c00650072002000730065006e006500720065002egt PTB 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName 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S857770133page 33

MARKS DO NOT WRITE IN

THIS MARGIN

13 The Nobel Prize-winning physicist Richard Feynman once stated lsquothings on a small scale behave nothing like things on a large scalersquo

Using your knowledge of physics comment on his statement

[Turn over

3

S857770134page 34

MARKS DO NOT WRITE IN

THIS MARGIN

14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

MARKS DO NOT WRITE IN

THIS MARGIN

14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

S857770139page 39

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN ltFEFF004200720075006700200069006e0064007300740069006c006c0069006e006700650072006e0065002000740069006c0020006100740020006f007000720065007400740065002000410064006f006200650020005000440046002d0064006f006b0075006d0065006e007400650072002c0020006400650072002000620065006400730074002000650067006e006500720020007300690067002000740069006c002000700072006500700072006500730073002d007500640073006b007200690076006e0069006e00670020006100660020006800f8006a0020006b00760061006c0069007400650074002e0020004400650020006f007000720065007400740065006400650020005000440046002d0064006f006b0075006d0065006e0074006500720020006b0061006e002000e50062006e00650073002000690020004100630072006f00620061007400200065006c006c006500720020004100630072006f006200610074002000520065006100640065007200200035002e00300020006f00670020006e0079006500720065002egt DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

S857770134page 34

MARKS DO NOT WRITE IN

THIS MARGIN

14 A student carries out a series of experiments to investigate properties of capacitors in AC circuits

(a)  The student connects a 5∙0 μF capacitor to an AC supply of EMF 15 Vrms and negligible internal resistance as shown in Figure 14A

15 Vrms

50 microF

A

Figure 14A

The frequency of the AC supply is 65 Hz

(i) Calculate the reactance of the capacitor

Space for working and answer

(ii) Calculate the current in the circuit

Space for working and answer

3

3

S857770135page 35

MARKS DO NOT WRITE IN

THIS MARGIN

14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

S857770139page 39

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 ITA 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PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

S857770135page 35

MARKS DO NOT WRITE IN

THIS MARGIN

14 (continued)

(b) The student uses the following circuit to determine the capacitance of a second capacitor

A

V

signal generator

Figure 14B

The student obtains the following data

Xc (Ω) f (Hz)

1∙60 times 106 10

6∙47 times 105 40

2∙99 times 105 100

1∙52 times 105 200

6∙35 times 104 500

3∙18 times 104 1000

(i) On the square-ruled paper on page 37 plot a graph of Xc against1f

(ii) Use your graph to determine the capacitance of this capacitor

Space for working and answer

[END OF SPECIMEN QUESTION PAPER]

3

3

S857770136page 36

S857770137page 37

S857770138page 38

S857770139page 39

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

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ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN ltFEFF004200720075006700200069006e0064007300740069006c006c0069006e006700650072006e0065002000740069006c0020006100740020006f007000720065007400740065002000410064006f006200650020005000440046002d0064006f006b0075006d0065006e007400650072002c0020006400650072002000620065006400730074002000650067006e006500720020007300690067002000740069006c002000700072006500700072006500730073002d007500640073006b007200690076006e0069006e00670020006100660020006800f8006a0020006b00760061006c0069007400650074002e0020004400650020006f007000720065007400740065006400650020005000440046002d0064006f006b0075006d0065006e0074006500720020006b0061006e002000e50062006e00650073002000690020004100630072006f00620061007400200065006c006c006500720020004100630072006f006200610074002000520065006100640065007200200035002e00300020006f00670020006e0079006500720065002egt DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

S857770136page 36

S857770137page 37

S857770138page 38

S857770139page 39

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU ltFEFF00560065007200770065006e00640065006e0020005300690065002000640069006500730065002000450069006e007300740065006c006c0075006e00670065006e0020007a0075006d002000450072007300740065006c006c0065006e00200076006f006e002000410064006f006200650020005000440046002d0044006f006b0075006d0065006e00740065006e002c00200076006f006e002000640065006e0065006e002000530069006500200068006f006300680077006500720074006900670065002000500072006500700072006500730073002d0044007200750063006b0065002000650072007a0065007500670065006e0020006d00f60063006800740065006e002e002000450072007300740065006c006c007400650020005000440046002d0044006f006b0075006d0065006e007400650020006b00f6006e006e0065006e0020006d006900740020004100630072006f00620061007400200075006e0064002000410064006f00620065002000520065006100640065007200200035002e00300020006f0064006500720020006800f600680065007200200067006500f600660066006e00650074002000770065007200640065006e002egt ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

S857770137page 37

S857770138page 38

S857770139page 39

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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S857770138page 38

S857770139page 39

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA ltFEFF005500740069006c0069007a007a006100720065002000710075006500730074006500200069006d0070006f007300740061007a0069006f006e00690020007000650072002000630072006500610072006500200064006f00630075006d0065006e00740069002000410064006f00620065002000500044004600200070006900f900200061006400610074007400690020006100200075006e00610020007000720065007300740061006d0070006100200064006900200061006c007400610020007100750061006c0069007400e0002e0020004900200064006f00630075006d0065006e007400690020005000440046002000630072006500610074006900200070006f00730073006f006e006f0020006500730073006500720065002000610070006500720074006900200063006f006e0020004100630072006f00620061007400200065002000410064006f00620065002000520065006100640065007200200035002e003000200065002000760065007200730069006f006e006900200073007500630063006500730073006900760065002egt JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ 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S857770139page 39

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

S857770140page 40

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA ltFEFF005500740069006c0069007a007a006100720065002000710075006500730074006500200069006d0070006f007300740061007a0069006f006e00690020007000650072002000630072006500610072006500200064006f00630075006d0065006e00740069002000410064006f00620065002000500044004600200070006900f900200061006400610074007400690020006100200075006e00610020007000720065007300740061006d0070006100200064006900200061006c007400610020007100750061006c0069007400e0002e0020004900200064006f00630075006d0065006e007400690020005000440046002000630072006500610074006900200070006f00730073006f006e006f0020006500730073006500720065002000610070006500720074006900200063006f006e0020004100630072006f00620061007400200065002000410064006f00620065002000520065006100640065007200200035002e003000200065002000760065007200730069006f006e006900200073007500630063006500730073006900760065002egt JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB ltFEFF005500740069006c0069007a006500200065007300730061007300200063006f006e00660069006700750072006100e700f50065007300200064006500200066006f0072006d00610020006100200063007200690061007200200064006f00630075006d0065006e0074006f0073002000410064006f0062006500200050004400460020006d00610069007300200061006400650071007500610064006f00730020007000610072006100200070007200e9002d0069006d0070007200650073007300f50065007300200064006500200061006c007400610020007100750061006c00690064006100640065002e0020004f007300200064006f00630075006d0065006e0074006f00730020005000440046002000630072006900610064006f007300200070006f00640065006d0020007300650072002000610062006500720074006f007300200063006f006d0020006f0020004100630072006f006200610074002000650020006f002000410064006f00620065002000520065006100640065007200200035002e0030002000650020007600650072007300f50065007300200070006f00730074006500720069006f007200650073002egt SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

S857770140page 40

MARKS DO NOT WRITE IN

THIS MARGIN

ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Acknowledgement of copyrightQuestion 1 Calvin ChanShutterstockcom

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 DEU 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

AH

S8577711copy

NationalQualicationsSPECIMEN ONLY

S8577711 PhysicsRelationships sheet

Date mdash Not applicable

Duration mdash 3 hours

page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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page 02

Relationships required for Physics Advanced Higher

dsvdt

=

2

2

dv d sadt dt

= =

v u at= +

212

s ut at= +

2 2 2v u as= +

dθωdt

=

2

2

dω d θαdt dt

= =

oω ω αt= +

212oθ ω t αt= +

2 2 2oω ω αθ= +

s rθ=

v rω=

ta rα=

22

rva rωr

= =

22mvF mrω

r= =

2L mvr mr ω= =

L Iω=

GMVr

= minus

2 44L πr σT=

2

2Schwarzschild

GMrc

=

E hf=

hλp

=

2

nhmvrπ

=

4xhx pπ

Δ Δ ge

4

hE tπ

Δ Δ ge

F qvB=

2ω πf=

2πωT

=

PGMmE Vmr

= = minus

( )k rotationalE Iω= 212

τ Fr=

τ Iα=

I mr=sum 2

( ) ( )P k translational k rotationalE E E= +

GMm mv πF mrω mrr r T

= = = =

222

2

2

escGMvr

=2

Lbπd

= 24

P σTA= 4

mvFr

=2

GMmFr

= 2

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF00560065007200770065006e00640065006e0020005300690065002000640069006500730065002000450069006e007300740065006c006c0075006e00670065006e0020007a0075006d002000450072007300740065006c006c0065006e00200076006f006e002000410064006f006200650020005000440046002d0044006f006b0075006d0065006e00740065006e002c00200076006f006e002000640065006e0065006e002000530069006500200068006f006300680077006500720074006900670065002000500072006500700072006500730073002d0044007200750063006b0065002000650072007a0065007500670065006e0020006d00f60063006800740065006e002e002000450072007300740065006c006c007400650020005000440046002d0044006f006b0075006d0065006e007400650020006b00f6006e006e0065006e0020006d006900740020004100630072006f00620061007400200075006e0064002000410064006f00620065002000520065006100640065007200200035002e00300020006f0064006500720020006800f600680065007200200067006500f600660066006e00650074002000770065007200640065006e002egt ESP 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 FRA ltFEFF005500740069006c006900730065007a00200063006500730020006f007000740069006f006e00730020006100660069006e00200064006500200063007200e900650072002000640065007300200064006f00630075006d0065006e00740073002000410064006f00620065002000500044004600200070006f0075007200200075006e00650020007100750061006c0069007400e90020006400270069006d007000720065007300730069006f006e00200070007200e9007000720065007300730065002e0020004c0065007300200064006f00630075006d0065006e00740073002000500044004600200063007200e900e90073002000700065007500760065006e0074002000ea0074007200650020006f007500760065007200740073002000640061006e00730020004100630072006f006200610074002c002000610069006e00730069002000710075002700410064006f00620065002000520065006100640065007200200035002e0030002000650074002000760065007200730069006f006e007300200075006c007400e90072006900650075007200650073002egt ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 03

cos sinory A ωt y A ωt= =

2 212PE mω y=

2E kA=

2πxλ

=φ

2

λlxd

Δ =

4

λdn

=

λDxd

Δ =

tan Pn i=

1 224 o

QQFπε r

=

24 o

QEπε r

=

4 o

QVπε r

=

F QE=

V Ed=

sinF IlB θ=

2oμ IBπr

=

1

o o

cε μ

=

CVXI

=

12CX πfC

=

212

E LI=

LVXI

=

2LX πfL=

2 2 2W X Y ZW X Y Z

Δ Δ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2W X Y ZΔ = Δ + Δ + Δ

22

2

d ya ω ydt

= = minus

F ky= minus

πω πfT

= =2

2

( )v ω A y= plusmn minus2 2

( )kE mω A y= minus2 2 212

sin xy A π ftλ

= minus

2

opd n gpd= times

W QV=

kE mv= 212

F qvB=

τ RC=

Δ Δn

n

W WnW W

=

1or where 0 1 2

2 opd mλ m λ m = + =

dIε Ldt

= minus

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 DEU 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 ESP ltFEFF005500740069006c0069006300650020006500730074006100200063006f006e0066006900670075007200610063006900f3006e0020007000610072006100200063007200650061007200200064006f00630075006d0065006e0074006f00730020005000440046002000640065002000410064006f0062006500200061006400650063007500610064006f00730020007000610072006100200069006d0070007200650073006900f3006e0020007000720065002d0065006400690074006f007200690061006c00200064006500200061006c00740061002000630061006c0069006400610064002e002000530065002000700075006500640065006e00200061006200720069007200200064006f00630075006d0065006e0074006f00730020005000440046002000630072006500610064006f007300200063006f006e0020004100630072006f006200610074002c002000410064006f00620065002000520065006100640065007200200035002e003000200079002000760065007200730069006f006e0065007300200070006f00730074006500720069006f007200650073002egt FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE ltFEFF0041006e007600e4006e00640020006400650020006800e4007200200069006e0073007400e4006c006c006e0069006e006700610072006e00610020006f006d002000640075002000760069006c006c00200073006b006100700061002000410064006f006200650020005000440046002d0064006f006b0075006d0065006e007400200073006f006d002000e400720020006c00e4006d0070006c0069006700610020006600f60072002000700072006500700072006500730073002d007500740073006b00720069006600740020006d006500640020006800f600670020006b00760061006c0069007400650074002e002000200053006b006100700061006400650020005000440046002d0064006f006b0075006d0065006e00740020006b0061006e002000f600700070006e00610073002000690020004100630072006f0062006100740020006f00630068002000410064006f00620065002000520065006100640065007200200035002e00300020006f00630068002000730065006e006100720065002egt ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 04

d vt=

s vt=

v u at= +

212

s ut at= +

2 2 2v u as= +

( )12

s u v t= +

W mg=

F ma=

WE Fd=

PE mgh=

212KE mv=

EPt

=

p mv=

Ft mv mu= minus

2

MmF Gr

=

( )21

ttvc

prime =minus

( )21 vl l cprime = minus

o ss

vf fv v

⎛ ⎞= ⎜ ⎟plusmn⎝ ⎠

observed rest

rest

λ λzλ

minus=

vzc

=

0v H d=

W QV=

2E mc=

E hf=

0KE hf hf= minus

2 1E E hfminus =

1Tf

=

v fλ=

sind θ mλ=

sinsin

1

2

θnθ

=

sinsin

1 1 1

2 2 2

θ λ vθ λ v

= =

sin 1cθ n

=

2

kId

=

PIA

=

1path difference or where 0 1 22

mλ m λ m⎛ ⎞= +⎜ ⎟⎝ ⎠

=

max value min valuerandom uncertainty

number of valuesminus=

2peak rmsV V=

2peak rmsI I=

Q It=

V IR=

22 VP IV I R

R= = =

1 2TR R R= + +

1 2

1 1 1

TR R R= + +

E V Ir= +

11

1 2S

RV V

R R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

1 1

2 2

V RV R

=

221 1 1

2 2 2QE QV CVC

= = =

QCV

=

page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ 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page 05

Additional relationships

Circle

Sphere

Trigonometry

Moment of inertia

point mass

rod about centre

rod about end

disc about centre

sphere about centre

Table of standard derivatives

( )f x ( )f xprime

sin ax cosa ax

cosax sina axminus

Table of standard integrals

( )f x ( )f x dxint

sin ax cos1 ax Caminus +

cosax sin1 ax Ca +

circumference 2πr=

2area πr=

2area 4πr=

343volume πr=

sin oppositehypotenuse

θ =

cos adjacenthypotenuse

θ =

tan oppositeadjacent

θ =

sin cos2 2 1θ θ+ =

2I mr=

2112I ml=

213I ml=

212I mr=

225I mr=

page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

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• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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page 06

Electron arrangements of elem

entsG

roup 1G

roup 2G

roup 3G

roup 4G

roup 5G

roup 6G

roup 7G

roup 0

(1)(18)

1H1

Hydrogen

KeyAtom

ic number

Symbol

Electron arrangement

Nam

e

2He2

Helium

(13)(14)

(15)(16)

(17)(2)

3Li21

Lithium

4Be22

Beryllium

5B23

Boron

6C24

Carbon

7N25

Nitrogen

8O26

Oxygen

9F27

Fluorine

10

Ne

28

Neon

11

Na

281

Sodium

12

Mg

282

Magnesium

Transition elements

13Al283

Aluminium

14Si284

Silicon

15P285

Phosphorus

16S286

Sulfur

17Cl287

Chlorine

18

Ar288

Argon(3)

(4)(5)

(6)(7)

(8)(9)

(10)(11)

(12)

19K2881

Potassium

20

Ca2882

Calcium

21Sc2892

Scandium

22Ti28102

Titanium

23V28112

Vanadium

24

Cr28131

Chromium

25

Mn

28132

Manganese

26

Fe28142

Iron

27

Co28152

Cobalt

28Ni

28162

Nickel

29

Cu28181

Copper

30

Zn28182

Zinc

31

Ga

28183

Gallium

32

Ge

28184

Germ

anium

33

As28185

Arsenic

34

Se28186

Selenium

35

Br28187

Bromine

36

Kr28188

Krypton

37

Rb281881

Rubidium

38Sr281882

Strontium

39Y281892

Yttrium

40Zr2818

102Zirconium

41

Nb

2818121

Niobium

42

Mo

2818131

Molybdenum

43Tc281813

2Technetium

44

Ru281815

1Ruthenium

45

Rh281816

1Rhodium

46

Pd2818

180Palladium

47

Ag2818

181Silver

48

Cd2818

182Cadm

ium

49In2818

183Indium

50

Sn2818

184Tin

51

Sb2818

185Antim

ony

52

Te2818

186Tellurium

53I2818

187Iodine

54

Xe2818

188Xenon

55

Cs281818

81Caesium

56

Ba281818

82Barium

57

La281818

92Lanthanum

72

Hf

281832102

Hafnium

73

Ta281832112Tantalum

74W281832

122Tungsten

75

Re281832

132Rhenium

76

Os

281832142

Osm

ium

77Ir281832

152Iridium

78Pt281832

171Platinum

79

Au281832181

Gold

80

Hg

281832182M

ercury

81Tl281832183Thallium

82

Pb281832184

Lead

83Bi281832185Bism

uth

84

Po281832186Polonium

85

At281832187Astatine

86

Rn281832188Radon

87Fr281832

1881Francium

88

Ra281832

1882Radium

89

Ac281832

1892Actinium

104

Rf281832

32102Rutherfordium

105

Db

28183232112D

ubnium

106

Sg281832

32122Seaborgium

107

Bh281832

32132Bohrium

108

Hs

28183232142H

assium

109

Mt

28183232152

Meitnerium

110

Ds

28183232171

Darmstadtium

111

Rg281832

32181Roentgenium

112

Cn281832

32182Copernicium

57

La28181892

Lanthanum

58

Ce28182082Cerium

59Pr281821

82Praseodym

ium

60

Nd

28182282

Neodym

ium

61

Pm281823

82Prom

ethium

62

Sm281824

82Sam

arium

63

Eu281825

82Europium

64

Gd

28182592

Gadolinium

65

Tb281827

82Terbium

66

Dy

28182882

Dysprosium

67

Ho

28182982

Holm

ium

68Er281830

82Erbium

69

Tm281831

82Thulium

70

Yb281832

82Ytterbium

71

Lu281832

92Lutetium

89

Ac281832

1892Actinium

90

Th281832

18102Thorium

91

Pa281832

2092Protactinium

92U281832

2192U

ranium

93

Np

2818322292

Neptunium

94

Pu281832

2482Plutonium

95

Am281832

2582Am

ericium

96

Cm281832

2592Curium

97

Bk281832

2782Berkelium

98Cf281832

2882Californium

99Es281832

2982Einsteinium

100

Fm281832

3082Ferm

ium

101

Md

2818323182

Mendelevium

102

No

2818323282

Nobelium

103

Lr281832

3292Law

rencium

Lanthanides

Actinides

AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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AH

copy

NationalQualicationsSPECIMEN ONLY

S8577701 Physics

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis If it is reproduced SQA should be clearly acknowledged as the source If it is to be used for any other purpose written permission must be obtained from permissionssqaorguk

Where the publication includes materials from sources other than SQA (ie secondary copyright) this material should only be reproduced for the purposes of examination or assessment If it needs to be reproduced for any other purpose it is the userrsquos responsibility to obtain the necessary copyright clearance

These marking instructions have been provided to show how SQA would mark this specimen question paper

page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN 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page 02

General marking principles for Physics Advanced Higher

Always assign marks for each candidate response in line with these marking principles the Physics general marking principles (GMPs) (httpwwwsqaorgukfiles_cccPhysicsgeneralmarkingprinciplespdf) and the detailed marking instructions for this assessment

(a) Always use positive marking This means candidates accumulate marks for the demonstration of relevant skills knowledge and understanding marks are not deducted from a maximum on the basis of errors or omissions

(b) If a specific candidate response does not seem to be covered by either the principles or detailed marking instructions and you are uncertain how to assess it you must seek guidance from your team leader

(c) Where a candidate incorrectly answers part of a question and carries the incorrect answer forward in the following part award marks if the incorrect answer has then been used correctly in the subsequent part or lsquofollow-onrsquo (GMP 17)

(d) Award full marks for a correct final answer (including units if required) on its own unless a numerical question specifically requires evidence of working to be shown eg in a lsquoshowrsquo question (GMP 1)

(e) Award marks where a diagram or sketch correctly conveys the response required by the question Clear and correct labels (or the use of standard symbols) are usually required for marks to be awarded (GMP 19)

(f) Award marks for knowledge of relevant relationships alone When a candidate writes down several relationships and does not select the correct one to continue with for example by substituting values do not award a mark (GMP 3)

(g) Award marks for the use of non-standard symbols where the symbols are defined and the relationship is correct or where the substitution shows that the relationship used is correct This must be clear and unambiguous (GMP 22)

(h) Do not award marks if a lsquomagic trianglersquo (eg) is the only statement in a candidatersquos

response To gain the mark the correct relationship must be stated for example V = IR or V

RI

= (GMP 6)

(i) In rounding to an expected number of significant figures award the mark for responses that have up to two figures more or one figure less than the number in the data with the fewest significant figures (GMP 10)

For example

Data in question is given to 3 significant figures

Correct final answer is 8middot16 J

Final answer 8middot2 J or 8middot158 J or 8middot1576 J ndash award the final mark

Final answer 8 J or 8middot15761 J ndash do not award the final mark

(Note the use of a recurrence dot eg 0middot6 would imply an infinite number of significant figures and would therefore not be acceptable)

page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 ESP 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ltFEFF005500740069006c0069007a007a006100720065002000710075006500730074006500200069006d0070006f007300740061007a0069006f006e00690020007000650072002000630072006500610072006500200064006f00630075006d0065006e00740069002000410064006f00620065002000500044004600200070006900f900200061006400610074007400690020006100200075006e00610020007000720065007300740061006d0070006100200064006900200061006c007400610020007100750061006c0069007400e0002e0020004900200064006f00630075006d0065006e007400690020005000440046002000630072006500610074006900200070006f00730073006f006e006f0020006500730073006500720065002000610070006500720074006900200063006f006e0020004100630072006f00620061007400200065002000410064006f00620065002000520065006100640065007200200035002e003000200065002000760065007200730069006f006e006900200073007500630063006500730073006900760065002egt JPN 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geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR ltFEFF004200720075006b00200064006900730073006500200069006e006e007300740069006c006c0069006e00670065006e0065002000740069006c002000e50020006f0070007000720065007400740065002000410064006f006200650020005000440046002d0064006f006b0075006d0065006e00740065007200200073006f006d00200065007200200062006500730074002000650067006e0065007400200066006f00720020006600f80072007400720079006b006b0073007500740073006b00720069006600740020006100760020006800f800790020006b00760061006c0069007400650074002e0020005000440046002d0064006f006b0075006d0065006e00740065006e00650020006b0061006e002000e50070006e00650073002000690020004100630072006f00620061007400200065006c006c00650072002000410064006f00620065002000520065006100640065007200200035002e003000200065006c006c00650072002000730065006e006500720065002egt PTB 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 SVE 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page 03

(j) Award marks where candidates have incorrectly spelled technical terms provided that responses can be interpreted and understood without any doubt as to the meaning Where there is ambiguity do not award the mark Two specific examples of this would be when the candidate uses a term that might be interpreted as lsquoreflectionrsquo lsquorefractionrsquo or lsquodiffractionrsquo (for example lsquodefractionrsquo) or one that might be interpreted as either lsquofissionrsquo or lsquofusionrsquo (for example lsquofussionrsquo) (GMP 25)

(k) Only award marks for a valid response to the question asked Where candidates are asked to

bull identify name give or state they need only name or present in brief form

bull describe they must provide a statement or structure of characteristics andor features

bull explain they must relate cause and effect andor make relationships between things clear

bull determine or calculate they must determine a number from given facts figures or information

bull estimate they must determine an approximate value for something bull justify they must give reasons to support their suggestions or conclusions For example

this might be by identifying an appropriate relationship and the effect of changing variables

bull show that they must use physics [and mathematics] to prove something for example a given value ndash all steps including the stated answer must be shown

bull predict they must suggest what may happen based on available information bull suggest they must apply their knowledge and understanding of physics to a new situation

A number of responses are acceptable award marks for any suggestions that are supported by knowledge and understanding of physics

bull use their knowledge of physics or aspect of physics to comment on they must apply their skills knowledge and understanding to respond appropriately to the problemsituation presented (for example by making a statement of principle(s) involved andor a relationship or equation and applying these to respond to the problemsituation) Candidates are given credit for the breadth andor depth of their conceptual understanding

page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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page 04

(l) Marking in calculations Example question The current in a resistor is 1middot5 amperes when the potential difference across it is 7middot5 volts Calculate the resistance of the resistor (3 marks)

Example response Mark and comment

1 V = IR

7middot5 = 1middot5R

R = 5middot0 Ω

1 mark relationship

1 mark substitution

1 mark correct answer

2 5middot0 Ω 3 marks correct answer

3 5middot0 2 marks unit missing

4 4middot0 Ω 0 marks no evidence wrong answer

5 __ Ω 0 marks no working or final answer

6 ΩV

RI

= = =7middot54 0

1middot5middot 2 marks arithmetic error

7 ΩV

RI

= = 4 0middot 1 mark relationship only

8 __V

RI

= = Ω 1 mark relationship only

9 __ΩV

RI

= = =7middot51middot5

2 marks relationship and substitution no final answer

10 VR

I= = =7middot5

4 01middot5

middot 2 marks relationship and substitution wrong answer

11 ΩV

RI

= = =1middot55middot0

7middot5 1 mark relationship but wrong substitution

12 ΩV

RI

= = =755 0

1middot5middot 1 mark relationship but wrong substitution

13 ΩI

RV

= = =1middot55 0

7middot5middot 0 marks wrong relationship

14 V = IR

7middot5 = 1middot5 times R

R = 0middot2 Ω

2 marks relationship and substitution arithmetic error

15 V = IR

Ω1middot50 2

7middot5middot

IR

V= = =

1 mark relationship correct but wrong rearrangement of symbols

page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN 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geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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page 05

Marking instructions for each question

Question Expected response Max mark Additional guidance

1 (a)

( )

2

2

0middot135 1middot26

0middot135 2 1middot26

0middot135 2 15middot0 1middot26

5middot31ms

v t t

dva t

dt

a

a minus

= +

⎛ ⎞= = times +⎜ ⎟⎝ ⎠= times times +=

1

1

1

( )

( )

( )

3 Accept 5middot3 5middot310 5middot3100

(b)

( ) ( )

v t t

s v dt t t

s

s

= +

⎡ ⎤= = +⎣ ⎦

= times + times=

int

2

15middot015middot0 3 2

0 0

3 2

0middot135 1middot26

0middot045 0middot63

0middot045 15middot0 0middot063 15middot0

294 m

1

1

1

( )

( )

( )

3 Accept 290 293middot6 293middot63 Constant of integration method acceptable

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 ESP 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 FRA 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 ITA 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 JPN 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geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR ltFEFF004200720075006b00200064006900730073006500200069006e006e007300740069006c006c0069006e00670065006e0065002000740069006c002000e50020006f0070007000720065007400740065002000410064006f006200650020005000440046002d0064006f006b0075006d0065006e00740065007200200073006f006d00200065007200200062006500730074002000650067006e0065007400200066006f00720020006600f80072007400720079006b006b0073007500740073006b00720069006600740020006100760020006800f800790020006b00760061006c0069007400650074002e0020005000440046002d0064006f006b0075006d0065006e00740065006e00650020006b0061006e002000e50070006e00650073002000690020004100630072006f00620061007400200065006c006c00650072002000410064006f00620065002000520065006100640065007200200035002e003000200065006c006c00650072002000730065006e006500720065002egt PTB ltFEFF005500740069006c0069007a006500200065007300730061007300200063006f006e00660069006700750072006100e700f50065007300200064006500200066006f0072006d00610020006100200063007200690061007200200064006f00630075006d0065006e0074006f0073002000410064006f0062006500200050004400460020006d00610069007300200061006400650071007500610064006f00730020007000610072006100200070007200e9002d0069006d0070007200650073007300f50065007300200064006500200061006c007400610020007100750061006c00690064006100640065002e0020004f007300200064006f00630075006d0065006e0074006f00730020005000440046002000630072006900610064006f007300200070006f00640065006d0020007300650072002000610062006500720074006f007300200063006f006d0020006f0020004100630072006f006200610074002000650020006f002000410064006f00620065002000520065006100640065007200200035002e0030002000650020007600650072007300f50065007300200070006f00730074006500720069006f007200650073002egt SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 06

Question Expected response Max mark Additional guidance

2 (a)

π

θωt

ω

ω minus

=

times times=

= 1

1250 260

131 rads

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i) ω ωα

t

α

α minus

minus=

minus=

=

0

2

131 7middot50

12middot0

10middot3 rads

1

1

1

( )

( )

( )

3 Accept 10 10middot29 10middot292

(b) (ii)

( ) ( )

( ) ( )π

θ ω t αt

θ

= +

= times + times times

times + times times=

=

20

2

2

12

7middot50 12middot0 0middot5 10middot3 12middot0

7middot50 12middot0 0middot5 10middot3 12middot0revolutions

2revolutions 132

1

1

1

1

( )

( )

( )

( )

4 Or consistent with (b) (i) Accept 130 132middot4 132middot35

(c) F mrω

F

F

minus

== times times times=

2

2 21middot5 10 0middot28 131

72 N

1

1

1

( )

( )

( )

3 Accept 70 72middot1 72middot08

(d) (i) The drum exerts a centripetalcentral force on the clothing (1) No centripetalcentral force acting on water (1)

2

(ii)

1

(iii) Centripetal force decreases (1) as mass of wet clothing decreases (1)

2

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR ltFEFF004200720075006b00200064006900730073006500200069006e006e007300740069006c006c0069006e00670065006e0065002000740069006c002000e50020006f0070007000720065007400740065002000410064006f006200650020005000440046002d0064006f006b0075006d0065006e00740065007200200073006f006d00200065007200200062006500730074002000650067006e0065007400200066006f00720020006600f80072007400720079006b006b0073007500740073006b00720069006600740020006100760020006800f800790020006b00760061006c0069007400650074002e0020005000440046002d0064006f006b0075006d0065006e00740065006e00650020006b0061006e002000e50070006e00650073002000690020004100630072006f00620061007400200065006c006c00650072002000410064006f00620065002000520065006100640065007200200035002e003000200065006c006c00650072002000730065006e006500720065002egt PTB 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 07

Question Expected response Max mark Additional guidance

3 (a) I mr

I

I

=

= times times=

2

2

2

12

0middot5 6middot0 0middot5

0middot75 kgm

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (i)

( )Total

Total

I mr

I

I

I

== times

= times +

=

2

2

2

2

(For 2middot0 kg mass)

2middot0 0middot40

2middot0 0middot40 0middot75

1middot1 kgm

1

1

1

( )

( )

( )

3 Accept 1 1middot07 1middot070

(ii) I ω I ω

ωω minus

=times = times

=

1 1 2 2

2

12

0middot75 12 1middot1

8middot2 rads

1

1

1

( )

( )

( )

3 Or consistent with (b)(i) Accept 8 8middot18 8middot182

(iii) No external torque acts on system or 2 kg can be considered as a point mass

1

(c) The (final) angular velocity will be greater (1) The final moment of inertia is less than in (b)(ii) (1)

2 To be awarded the second mark the candidate must make reference to moment of inertia It is insufficient to say sphere rolls off

page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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page 08

Question Expected response Max mark Additional guidance

4 (a) (i)

π

π

π

E

E

GMmmrω

r

ωT

GM mm r

r T

rT

GM

=

=

=

=

22

2

2 2

3

2

4

2

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(ii) ( )π

6 5 3

11 24

3

6middot4 10 4middot0 102

6middot67 10 6middot0 10

5middot6 10 s

T

T

minus

times + times=times times times

= times

1

1

( )

( )

2 Accept 6 5middot57 5middot569

(b) (i)

( )

5

2

11 24

5 6 2

12

Value from graph 4middot15 10 (m)

6middot67 10 6middot0 10

4middot15 10 6middot4 10

8middot6 Nkg

GMmmg

r

g

ω

minus

minus

= times

=

times times times=times + times

=

1

1

1

1

( )

( )

( )

( )

4 Accept 8 8middot62 8middot617

(ii) Less atmospheric dragfriction or will reduce running costs

1

(c) The gravitational field is smaller at the ISS (compared to Earth) (1) The clocks on ISS will run faster (than those on Earth) (1)

2

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU ltFEFF00560065007200770065006e00640065006e0020005300690065002000640069006500730065002000450069006e007300740065006c006c0075006e00670065006e0020007a0075006d002000450072007300740065006c006c0065006e00200076006f006e002000410064006f006200650020005000440046002d0044006f006b0075006d0065006e00740065006e002c00200076006f006e002000640065006e0065006e002000530069006500200068006f006300680077006500720074006900670065002000500072006500700072006500730073002d0044007200750063006b0065002000650072007a0065007500670065006e0020006d00f60063006800740065006e002e002000450072007300740065006c006c007400650020005000440046002d0044006f006b0075006d0065006e007400650020006b00f6006e006e0065006e0020006d006900740020004100630072006f00620061007400200075006e0064002000410064006f00620065002000520065006100640065007200200035002e00300020006f0064006500720020006800f600680065007200200067006500f600660066006e00650074002000770065007200640065006e002egt ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 09

Question Expected response Max mark Additional guidance

5 (a)

(Betelgeuse will look) red-orange 1

(b) (i)A π

π ( ) ( )L r σT

L

L

minus

== times times times times times times times= times

2 4

10 2 8 4 4

31

4

4 5middot49 10 5middot67 10 1middot2 10

4middot45 10 W

1

1

1

( )

( )

( )

3 Accept 45 4453 4453

(i)B Rigelstars behave as black body(ies) 1

(ii) Δ

Δ

Δ Δ Δ Δ

Δ

Δ

r

T

W X Y Z

W X Y Z

L

= times

= times

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= times times + times times⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= times

2 2 2

2 2

31

0middot05 100

5middot49

0middot05 100

1middot20

0middot05 0middot05Total 100 2 100 4

5middot49 1middot20

1 10 W

1 both

1

1

1

( ) ( )

( )

( )

( )

4 Or consistent with (b)(i)A

(c)

π

π ( )

2

31

8 2

8 2

4

4middot45 10

4 773 365middot25 24 60 60 3middot00 10

6middot61 10 Wm

Lb

d

b

b minus minus

=

times=times times times times times times times times

= times

1

1 1

1

( )

( )

( )

4 Or consistent with (b)(i)A Accept 6middot6 6middot612 6middot6122

Independent mark for ly to m conversion (line 2) Accept the use of 365 or 365middot24

(d) Hydrogen fusion has stopped (in the core) (1) The core shrinks (1)

2

(e) Thermal pressure becomes greater than gravitational force causing Betelgeuse to expand (1) As Betelgeuse expands the thermal pressure and gravitational force eventually return to equilibrium (1)

2

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR ltFEFF004200720075006b00200064006900730073006500200069006e006e007300740069006c006c0069006e00670065006e0065002000740069006c002000e50020006f0070007000720065007400740065002000410064006f006200650020005000440046002d0064006f006b0075006d0065006e00740065007200200073006f006d00200065007200200062006500730074002000650067006e0065007400200066006f00720020006600f80072007400720079006b006b0073007500740073006b00720069006600740020006100760020006800f800790020006b00760061006c0069007400650074002e0020005000440046002d0064006f006b0075006d0065006e00740065006e00650020006b0061006e002000e50070006e00650073002000690020004100630072006f00620061007400200065006c006c00650072002000410064006f00620065002000520065006100640065007200200035002e003000200065006c006c00650072002000730065006e006500720065002egt PTB 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PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 10

Question Expected response Max mark Additional guidance

6 (a)

Electrostatic force (between the nucleusproton and the electron)

1

(b) (i)

π

π

π

π

21 2

20

2 2

20

22

0

2

0

4

4

12 8

8

k

Q Q mv

ε r r

e mv

ε r r

emv

ε r

eE

ε r

=

=

=

=

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(b) (ii)

π

( )π

k

k

k

eE

ε r

E

E

minus

minus minus

minus

=

times=times times times times times

= times

2

0

19 2

12 9

19

8

1middot60 10

8 8middot85 10 0middot21 10

5middot5 10 J

1

1

( )

( )

2 Accept 5 548 5483

(c)

π

π

nhmvr

n

n

minusminus

=

times timestimes =

=

3434

2

6middot63 104middot22 10

24 (must be an integer)

1

1

1

( )

( )

( )

3

(d) (i)

min

Δ Δπ

Δπ

Δ

x

hx p

x

x

minusminus

minus

ge

timestimes times ge

= times

3426

9

4

6middot63 101middot5 10

4

3middot5 10 m

1

1

1

( )

( )

( )

3 Accept 4 3middot52 3middot517

(ii) Δ Δ

πh

E t ge4

1( )

If Δt is small then ΔE is large (1) Therefore the largest possible energy of the electron may be big enough to overcome the repulsion and cross the gap (1) or

Δ Δπx

hx p ge

41( )

If the momentum is measured with a small uncertainty the uncertainty in the position of the electron is large enough (1) for the electron to exist on the other side of the gap(1)

3

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 11

Question Expected response Max mark Additional guidance

7 (a)

(Component of) velocity perpendicular to the magnetic field (lines) results in circular motion or central force (1) (Component of) velocity parallel to the magnetic field (lines) results in constant velocity or no force acting parallel to the magnetic field (1)

2

(b)

sin( sin )

sin

mvqv

r

v

r

r

minus

minus minus

=

= times timestimes times times times

= times times times times times times= times

2

7

27 7 2

19 7 6

3

B

2middot6 10 50

1middot673 10 2middot6 10 50

1middot6 10 2middot6 10 50 58 10

3middot6 10 m

1 1

1

1

1

( + )

( )

( )

( )

5 Award 1 mark for both relationships Award 1 mark for equating the relationships Accept 4 359 3591

(c) Any two statements

The anti proton path will turn in the opposite direction The anti proton helix pitch will be greater Radius of curvature will be smaller

2

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 DEU 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 ESP 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 FRA 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ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn 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PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 12

Question Expected response Max mark Additional guidance

8 (a) (i) Acceleration is proportional to displacement (from a fixed point) and is always directed to (that) fixed point or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point

1 Accept a ky= minus or F k y= minus

(a) (ii) T

πωTπω

ω minus

=

=

times=

= 1

From graph 5middot0 (s)

2

2

5middot0

1middot3 rads

1

1

1

1

( )

( )

( )

( )

4 Accept 1 126 1257

(a) (iii)

( ) ( )( )

a ω x

a

a minus

= minus= minus times minus= minus

2

2

2

1middot3 4middot0

6middot8 ms

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 7 676 6760

(a) (iv) Sine curve (1) Period of oscillation 5middot0 s (1) Maximum speed 5middot2 m sminus1 (1)

3 Award a maximum of 2 marks if the labels units are missing

(a) (v) p

p

p

E mω y

E

E

=

= times times times

= times

2 2

2 2

3

12

0middot5 85 1middot3 4middot0

1middot1 10 J

1

1

1

( )

( )

( )

3 Or consistent with (a)(ii) Accept 1 115 1149

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP ltFEFF005500740069006c0069006300650020006500730074006100200063006f006e0066006900670075007200610063006900f3006e0020007000610072006100200063007200650061007200200064006f00630075006d0065006e0074006f00730020005000440046002000640065002000410064006f0062006500200061006400650063007500610064006f00730020007000610072006100200069006d0070007200650073006900f3006e0020007000720065002d0065006400690074006f007200690061006c00200064006500200061006c00740061002000630061006c0069006400610064002e002000530065002000700075006500640065006e00200061006200720069007200200064006f00630075006d0065006e0074006f00730020005000440046002000630072006500610064006f007300200063006f006e0020004100630072006f006200610074002c002000410064006f00620065002000520065006100640065007200200035002e003000200079002000760065007200730069006f006e0065007300200070006f00730074006500720069006f007200650073002egt FRA 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 ITA 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 JPN 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geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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ltFEFF004b00e40079007400e40020006e00e40069007400e4002000610073006500740075006b007300690061002c0020006b0075006e0020006c0075006f00740020006c00e400680069006e006e00e4002000760061006100740069007600610061006e0020007000610069006e006100740075006b00730065006e002000760061006c006d0069007300740065006c00750074007900f6006800f6006e00200073006f00700069007600690061002000410064006f0062006500200050004400460020002d0064006f006b0075006d0065006e007400740065006a0061002e0020004c0075006f0064007500740020005000440046002d0064006f006b0075006d0065006e00740069007400200076006f0069006400610061006e0020006100760061007400610020004100630072006f0062006100740069006c006c00610020006a0061002000410064006f00620065002000520065006100640065007200200035002e0030003a006c006c00610020006a006100200075007500640065006d006d0069006c006c0061002egt SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 13

Question Expected response Max mark Additional guidance

8 (b) Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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 ESP 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 FRA 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 ITA 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 JPN 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geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO ltFEFF004b00e40079007400e40020006e00e40069007400e4002000610073006500740075006b007300690061002c0020006b0075006e0020006c0075006f00740020006c00e400680069006e006e00e4002000760061006100740069007600610061006e0020007000610069006e006100740075006b00730065006e002000760061006c006d0069007300740065006c00750074007900f6006800f6006e00200073006f00700069007600690061002000410064006f0062006500200050004400460020002d0064006f006b0075006d0065006e007400740065006a0061002e0020004c0075006f0064007500740020005000440046002d0064006f006b0075006d0065006e00740069007400200076006f0069006400610061006e0020006100760061007400610020004100630072006f0062006100740069006c006c00610020006a0061002000410064006f00620065002000520065006100640065007200200035002e0030003a006c006c00610020006a006100200075007500640065006d006d0069006c006c0061002egt SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 14

Question Expected response Max mark Additional guidance

9 (a) Reflected wave interferes with transmitted wave (to produce points of destructive and constructive interference)

1

(b) Antinode (constructive) high energy so melted spots (1) Node (destructive) low energy so no melting(1)

2

(c) λ

λ

times =

=

14 0middot264

20middot132 m

1

1

( )

( )

2 Accept 013 01320 013200

(d) 9

8 1

2middot45 10 0middot132

3middot23 10 ms

v fλv

v minus

== times times= times

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(e) The range of the results is small so the results are precise (1) The difference between the mean value of the results and the accepted value of c is larger than the range so the results are not accurate (1)

2

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF00560065007200770065006e00640065006e0020005300690065002000640069006500730065002000450069006e007300740065006c006c0075006e00670065006e0020007a0075006d002000450072007300740065006c006c0065006e00200076006f006e002000410064006f006200650020005000440046002d0044006f006b0075006d0065006e00740065006e002c00200076006f006e002000640065006e0065006e002000530069006500200068006f006300680077006500720074006900670065002000500072006500700072006500730073002d0044007200750063006b0065002000650072007a0065007500670065006e0020006d00f60063006800740065006e002e002000450072007300740065006c006c007400650020005000440046002d0044006f006b0075006d0065006e007400650020006b00f6006e006e0065006e0020006d006900740020004100630072006f00620061007400200075006e0064002000410064006f00620065002000520065006100640065007200200035002e00300020006f0064006500720020006800f600680065007200200067006500f600660066006e00650074002000770065007200640065006e002egt ESP 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 FRA ltFEFF005500740069006c006900730065007a00200063006500730020006f007000740069006f006e00730020006100660069006e00200064006500200063007200e900650072002000640065007300200064006f00630075006d0065006e00740073002000410064006f00620065002000500044004600200070006f0075007200200075006e00650020007100750061006c0069007400e90020006400270069006d007000720065007300730069006f006e00200070007200e9007000720065007300730065002e0020004c0065007300200064006f00630075006d0065006e00740073002000500044004600200063007200e900e90073002000700065007500760065006e0074002000ea0074007200650020006f007500760065007200740073002000640061006e00730020004100630072006f006200610074002c002000610069006e00730069002000710075002700410064006f00620065002000520065006100640065007200200035002e0030002000650074002000760065007200730069006f006e007300200075006c007400e90072006900650075007200650073002egt ITA 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ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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(Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 15

Question Expected response Max mark Additional guidance

10 (a) A series of bright and dark spots 1 Accept fringes

(b) hλp

λ

λ

minus

minus

minus

minus minus

=

times=times times

= times

34

24

12

12 9

6middot63 10

1middot20 10 220

2middot5 10 (m)

Estimate in the range 10 to 10 metres

1

1

1

1

( )

( )

( )

( )

4 Statement of value of slit separation must be distinct from value of λ

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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PreserveOverprintSettings true StartPage 1 SubsetFonts false TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true 3Of9Barcode Goudy Goudy-Bold Goudy-BoldItalic Goudy-Italic Helvetica Helvetica-Bold Helvetica-BoldOblique ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt CHS ltFEFF4f7f75288fd94e9b8bbe5b9a521b5efa7684002000410064006f006200650020005000440046002065876863900275284e8e9ad88d2891cf76845370524d53705237300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c676562535f00521b5efa768400200050004400460020658768633002gt CHT ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO ltFEFF004b00e40079007400e40020006e00e40069007400e4002000610073006500740075006b007300690061002c0020006b0075006e0020006c0075006f00740020006c00e400680069006e006e00e4002000760061006100740069007600610061006e0020007000610069006e006100740075006b00730065006e002000760061006c006d0069007300740065006c00750074007900f6006800f6006e00200073006f00700069007600690061002000410064006f0062006500200050004400460020002d0064006f006b0075006d0065006e007400740065006a0061002e0020004c0075006f0064007500740020005000440046002d0064006f006b0075006d0065006e00740069007400200076006f0069006400610061006e0020006100760061007400610020004100630072006f0062006100740069006c006c00610020006a0061002000410064006f00620065002000520065006100640065007200200035002e0030003a006c006c00610020006a006100200075007500640065006d006d0069006c006c0061002egt SVE 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 16

Question Expected response Max mark Additional guidance

11 (a) 3 1

(b) c

ε μ

c

c

minus minus

minus

=

=times times times

= times

0 0

12 6

8 1

1

1

8middot93 10 1middot32 10

2middot91 10 ms

1

1

1

( )

( )

( )

3

(c) (i)

μ

μ

ε

μ

μ

minus

minus

minus

minus

times= timestimes

=

times= timestimes

=

8

0 6

0

14

0 12

0

0

5 10uncert in 100

1middot32 10

uncert in 3middot8

7 10uncert in 100

8middot93 10

uncert in 0middot8

Uncertainty in more significant

1

1

1

( )

( )

( )

3

(ii)

ε μ

ε μminus

= times times times

= times

8

0 0

6 1

0 0

3middot81 1uncert in 2middot91 10

2 100

1uncert in 6 10 ms

1

1

( )

( )

2

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt CHS ltFEFF4f7f75288fd94e9b8bbe5b9a521b5efa7684002000410064006f006200650020005000440046002065876863900275284e8e9ad88d2891cf76845370524d53705237300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c676562535f00521b5efa768400200050004400460020658768633002gt CHT ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE 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PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 17

Question Expected response Max mark

Additional guidance

12 (a) (i) Force acting per unit positive charge 1

(ii)

π π

π ( )

π ( )

Q Q

ε r ε r

Q

Q

minus

minus

minus

minus

=

minus times =times times times times

times times times times= minus times

1 22 2

0 0

9

12 2

212 2

92

4 4

8middot0 10

4 8middot85 10 0middot4

4 8middot85 10 0middot2

2middot0 10 C

1

1

( )

( )

2 SHOW question Award a maximum of 1 mark if final answer is not shown

(iii)

π

π

π

π

π

QV

ε r

V

V

minus

minus

minus

minus

minus

minus

minus

minus

=

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

10

9

1 12

9

2 12

9

12

9

12

4

8middot0 10

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

8middot0 10Potential at X

4 8middot85 10 0middot40

2middot0 10

4 8middot85 10 0middot20

Potential at X 270 V

1

1

1

1

1

( )

( )

( )

( )

( )

5 Accept 300 269middot8

(b) (i)

π

π

π

π

V

V

minus

minus

minus

minus

minus

minus

minus

minus

minus times=times times times times

minus times=times times times times

minus times=times times times times

minus times+times times times times

= minus

9

1 12

9

2 12

9

12

9

12

8middot0 10

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

8middot0 10Potential at P

4 8middot85 10 0middot50

2middot0 10

4 8middot85 10 0middot50

Potential at P 180 V

1

1

1

( )

( )

( )

3 Accept

200 179middot8

(ii) Potential difference = minus180 minus (minus270) = 90 (V) (1) E QV

E

E

minus

minus

== times times= times

9

8

1middot0 10 90

9middot0 10 J

1

1

1

( )

( )

( )

4 Or consistent with a(iii)(b)(i)

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR ltFEFF004200720075006b00200064006900730073006500200069006e006e007300740069006c006c0069006e00670065006e0065002000740069006c002000e50020006f0070007000720065007400740065002000410064006f006200650020005000440046002d0064006f006b0075006d0065006e00740065007200200073006f006d00200065007200200062006500730074002000650067006e0065007400200066006f00720020006600f80072007400720079006b006b0073007500740073006b00720069006600740020006100760020006800f800790020006b00760061006c0069007400650074002e0020005000440046002d0064006f006b0075006d0065006e00740065006e00650020006b0061006e002000e50070006e00650073002000690020004100630072006f00620061007400200065006c006c00650072002000410064006f00620065002000520065006100640065007200200035002e003000200065006c006c00650072002000730065006e006500720065002egt PTB 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 SUO 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 SVE ltFEFF0041006e007600e4006e00640020006400650020006800e4007200200069006e0073007400e4006c006c006e0069006e006700610072006e00610020006f006d002000640075002000760069006c006c00200073006b006100700061002000410064006f006200650020005000440046002d0064006f006b0075006d0065006e007400200073006f006d002000e400720020006c00e4006d0070006c0069006700610020006600f60072002000700072006500700072006500730073002d007500740073006b00720069006600740020006d006500640020006800f600670020006b00760061006c0069007400650074002e002000200053006b006100700061006400650020005000440046002d0064006f006b0075006d0065006e00740020006b0061006e002000f600700070006e00610073002000690020004100630072006f0062006100740020006f00630068002000410064006f00620065002000520065006100640065007200200035002e00300020006f00630068002000730065006e006100720065002egt ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 18

Question Expected response Max mark Additional guidance

13 Award 3 marks where the candidate has demonstrated a good understanding of the physics involved They show a good comprehension of the physics of the situation and provide a logically correct answer to the question posed This type of response might include a statement of the principles involved a relationship or an equation and the application of these to respond to the problem The answer does not need to be lsquoexcellentrsquo or lsquocompletersquo for the candidate to gain full marks Award 2 marks where the candidate has demonstrated a reasonable understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood the problem Award 1 mark where the candidate has demonstrated a limited understanding of the physics involved They make some statement(s) that are relevant to the situation showing that they have understood at least a little of the physics within the problem Award 0 marks where the candidate has not demonstrated an understanding of the physics involved There is no evidence that they have recognised the area of physics involved or they have not given any statement of a relevant physics principle Award this mark also if the candidate merely restates the physics given in the question

3 Candidates may use a variety of physics arguments to answer this question Award marks based on candidates demonstrating overall good reasonable limited or no understanding

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB 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 SUO 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 SVE ltFEFF0041006e007600e4006e00640020006400650020006800e4007200200069006e0073007400e4006c006c006e0069006e006700610072006e00610020006f006d002000640075002000760069006c006c00200073006b006100700061002000410064006f006200650020005000440046002d0064006f006b0075006d0065006e007400200073006f006d002000e400720020006c00e4006d0070006c0069006700610020006600f60072002000700072006500700072006500730073002d007500740073006b00720069006600740020006d006500640020006800f600670020006b00760061006c0069007400650074002e002000200053006b006100700061006400650020005000440046002d0064006f006b0075006d0065006e00740020006b0061006e002000f600700070006e00610073002000690020004100630072006f0062006100740020006f00630068002000410064006f00620065002000520065006100640065007200200035002e00300020006f00630068002000730065006e006100720065002egt ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice

page 19

Question Expected response Max mark

Additional guidance

14 (a) (i) π

πΩ

C

C

C

XfC

X

X

minus

=

=times times times times

=

6

1

2

1

2 65 5middot0 10

490

1

1

1

( )

( )

( )

3 Accept 500 489middot7

(ii) RMS

RMSC

RMS

RMS

VI

X

I

I minus

=

=

= times 2

15490

3middot1 10 A

1

1

1

( )

( )

( )

3 Or consistent with (a)(i) Accept 3 3middot06 3middot061

(b) (i) Linear scale on both axes (1) Acceptable axes labels (quantities and units) (1) Points plotted correctly and acceptable line of best fit (1)

3 If either scale is non-linear a maximum of 1 mark can be awarded Allow plusmn half box tolerance when plotting points

(ii) Gradient of best fit line (1)

Gradient =πC1

2

Or

πC =

times1

2 gradient (1)

Final value of C (1)

3 If a candidate uses data points that are not on their line of best fit a maximum of 1 mark can be awarded A representative gradient value of 3middot13 times 107 gives a capacitance of 5middot08 times 10minus9 F Final value of C must be consistent with candidatersquos value for gradient

[END OF SPECIMEN MARKING INSTRUCTIONS]

• AHSQPPhysics QP
• AHSQPPhysics Relationships Sheet
• AHSQPPhysics MI CVR
• AHSQPPhysics MI

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AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 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ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt DAN 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 DEU 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 ESP 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 FRA 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 PTB ltFEFF005500740069006c0069007a006500200065007300730061007300200063006f006e00660069006700750072006100e700f50065007300200064006500200066006f0072006d00610020006100200063007200690061007200200064006f00630075006d0065006e0074006f0073002000410064006f0062006500200050004400460020006d00610069007300200061006400650071007500610064006f00730020007000610072006100200070007200e9002d0069006d0070007200650073007300f50065007300200064006500200061006c007400610020007100750061006c00690064006100640065002e0020004f007300200064006f00630075006d0065006e0074006f00730020005000440046002000630072006900610064006f007300200070006f00640065006d0020007300650072002000610062006500720074006f007300200063006f006d0020006f0020004100630072006f006200610074002000650020006f002000410064006f00620065002000520065006100640065007200200035002e0030002000650020007600650072007300f50065007300200070006f00730074006500720069006f007200650073002egt SUO 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors NoConversion DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles true MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling LeaveUntagged UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice