142403 Mark Scheme Unit g484 the Newtonian World January

8/12/2019 142403 Mark Scheme Unit g484 the Newtonian World January

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Oxford Cambridge and RSA Examinations

GCE

Physics A

Advanced GCE

Unit G484: The Newtonian World

Mark Scheme for January 2013


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This mark scheme is published as an aid to teachers and students, to indicate the requirementsof the examination. It shows the basis on which marks were awarded by examiners. It does notindicate the details of the discussions which took place at an examiners meeting beforemarking commenced.

All examiners are instructed that alternative correct answers and unexpected approaches incandidates scripts must be given marks that fairly reflect the relevant knowledge and skills

demonstrated.

Mark schemes should be read in conjunction with the published question papers and the reporton the examination.

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Annotat ions

Annotat ion Meaning

Benefit of doubt given

Contradiction

Incorrect response

Error carried forward

Follow through

Not answered question

Benefit of doubt not given

Power of 10 error

Omission mark

Rounding error

Error in number of significant figures

Correct response

Arithmetic error

Wrong physics or equation


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G484 Mark Scheme

4

The abbreviations, annotations and conventions used in the detailed Mark Scheme are:

Annotat ion Meaning

/ Alternative and acceptable answers for the same marking point

(1) Separates marking points

reject Answers which are not worthy of credit

not Answers which are not worthy of credit

IGNORE Statements which are irrelevant

ALLOW Answers that can be accepted

( ) Words which are not essential to gain credit

__ Underlined words must be present in answer to score a mark

ecf Error carried forward

AW Alternative wording

ORA Or reverse argument


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G484 Mark Scheme

5

Subject-specific Marking Instructions

Note about significant figures:If the data given in a question is to 2 sf, then allow answers to 2 or more sf.If an answer is given to fewer than 2 sf, then penalise once only in the entire paper.

Any exception to this rule will be mentioned in the Guidance Column.

CATEGORISATION OF MARKS

The mark scheme categorise marks on the MACB scheme.

Bmarks: These are awarded as independent marks, which do not depend on other marks. For a B-marwhich it refers must be seen specifically in the candidates answers.

Mmarks: These are method marks upon whichA-marks (accuracy marks) later depend. For an M-markwhich it refers must be seen in the candidates answers. If a candidate fails to score a particula

dependentA-marks can be scored.

Cmarks: These are compensatory method marks which can be scored even if the points to which they the candidate, providing subsequent working gives evidence that they must have known it. Focarries a C-mark and the candidate does not write down the actual equation but does correct wcandidate knew the equation, then the C-mark is given.

Amarks: These are accuracy or answer marks, which either depend on an M-mark, or allow a C-mark t


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G484 Mark Scheme

6

Question Answer Marks

1 (a) Rate of change of momentum (of a body) is proportional /equal to the (net) force (acting on it)

and takes place in the direction of that force.

M1

A1

Al low: Force = change

Note: momentummmark.

Al lowthis mark if the M

(b) (i) (3 x 5) (7 x 2) = 10v

v= (15 14)/10

= 0.10 (m s-1)

to the right (AW)

C1

M1

A1

Signs must be correct fo

Al low1 sf answer

Not forwards/towards B

(ii) Impulse = 3(0.1 5)

( = - 14.7) = (-)15 (Ns)

to the left (AW)

M1

A1

Al low: ecf from (b)(i)

Ignoresign

Not backwards/towards

(iii) (Newtons 3rdlaw says)Force on B (due to A) is equal and opposite to force on A(due to B)

time (of contact) / t is same for bothANDImpulse = Ft

impulse on A is equal and opposite to impulse on B

M1

A1

A0

Al low: use of minus sig

Not : Action and reaction

Total 9


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G484 Mark Scheme

7


2 (a) (i)

r

vg

2

orr

GMv 2

grv

6102777 ..v

7400v (m s-1)

C1

C1

A1

Correct formula in anyAl low: use of afor g

Mark is for substitutionAny use of r = 800 km

Note:Answer to 3 sf is

(ii)

v

rT

2

7450

10272 6

.

T

= 6100 (s)

GM

rT

32

2 4

2411

362

2

10610676

10274

.

.T

T = 6100 (s)

C1

A1

Al low: possible ecf for

No ecffor use of r= 6Both score 0/2

Note:Answer to 3 sf uAnswer to 3 sf u

(b) (i) Number of orbits = 24 x 3600 (= 14.2)6080

14

B1 Al lowany correct metAl loworaNo ecf from a(ii)

(ii) Circumference = 2r

equatorial circumference = 2x 6400 = 13.4width of photograph 3000

(But each orbit crosses the equator twice hence)number of orbits = 6.7

This is fewer than 14 orbits so whole of Earths surface canbe photographed (AW)

C1

C1

A1

A0

Al low:

Circumference = 2rlength of equator cove

(But each orbit crossesmin width to be photog

< 3000 km so all of Eaone day


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G484 Mark Scheme

8


(c) suitable example: eg weather / spy / surveying / mapping /GPS

B1 Ignore TV / radio / com

Total 10


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G484 Mark Scheme

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3 (a) Force is proportional to the product of the masses andinversely proportional to the square of their separation(AW)

B1Al low:

2r

GmMF with

(b) (i)

2r

GmM

mgJ

11

282

1067.6

103.15.7

G

rgMJ

MJ = 1.9 x 1027 (kg )

C1

C1

A1

Al low: formula with m c

Al low: use of GM

T2

2 4

Note: mark is for substi

(ii)

2

2

M

A

A

M

r

r

g

g

210

28

1042

1031

57

.

.

.

Mg

gM= 2.2 x 10-4 (N kg-1)

C1

A1

Al low: use of2r

GMg J

21011

1042

9110676

.

..Mg

substitution

gM= 2.2 x 10-4 (N kg-

(iii) T2r3 OR T2/r3= constant ( = 42/GMJ)

3

8

310

2

2

1031

1042

27

.

.

.

MT

TM= 1.8 x104 (hours)

C1

C1

A1

Al low: possible ecf for MAl low: use of othercor

Note: mark is for substi

Note using times in sec2 marks

Total 9


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G484 Mark Scheme

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4 (a) Obtain a set of readings for:mass m, time period AND calculate frequency using f =

1/T.

Plot graphs of fagainst 1/m AND fagainst 1/mThe graph which is a straight line through the originprovides the correct relationship

Reference to one method of improving reliabilityeg counting more than 5 oscillations to find T orf

taking repeat measurements of T orf (and averagevalues)

time oscillations from equilibrium position

B1

B1

B1

B1

Not number of oscillatio

Al low: product method Select the relatio

Al low: plot ln fagainst lgradient= -1 the

1/m (B1)

(b) (i)3

1036

21

122

.max Afv

188.050

3

max

v

2

max

50

34.0

2

1

KE

3

max 101.7

KE (J)

C1

C1

A1

Note:mark is for substi

(ii)

32

2 103621122

.max Afa

990.max a (ms-2)

C1

A1

Note: mark is for correc


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G484 Mark Scheme

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(c) Reference to:kinetic energy (of masses and spring),gravitational potential energy (of mass and spring),elastic (potential) energy / strain energy of spring

KE: zero (at lowest point), increasing to max atequilibrium point, decreasing to zero (at highestpoint)

GPE: increases (as masses rise from lowest to highestpoint) (clearly worded ora)(AW)

strain / elastic energy:decreases (as masses rise from lowest to highestpoint)(clearly worded ora)(AW)

B1

B1

B1

B1

Note:mark to be awardNote: potentialmusscore this mark

Total 13


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G484 Mark Scheme

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5 (a) (i) n= number of moles (in sample)ANDN= number of atoms / molecules (in sample) B1

Note:bothdefinitions a

Not: particles / Avogadr

(ii) nor NANDT is constant (and R and k are constants)

for a fixed mass (of gas),pV = constant / p1/V

M1

A1

nRTor NkTis constant

(iii) Shows that Nm-2x m3= Nm B1 Al low: Use of base unit

(b) (i) Calculates px (1/V)-1at twopoints on the graph

values are the same pV= constant /p 1/V / nRT =constant

M1

A1

Expected values for pVmost pointsAl low: Correct calculati

Calculates intercorigin

and pV= consta

(ii) Number of moles in 0.050 kg = 0.05/0.016 ( = 3.125)

(K)

..

289

3181253

7500

nR

pVT

C)(16T

C1

C1

A1

Al low: possible ecf fromerror for use ofpV= 0.0

Note: Mark is for correc

Note: Allow full range oapproachese.g. use of a molecular

temperature of 305C

Total 9


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G484 Mark Scheme

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6 (a) (i) vibrate (about their fixed positions) B1 Al low: molecules vibrat

(ii) greater amplitude / greater frequency (of vibration) B1 Not:faster / more / bigg

(iii) Eitherinternal energy increases

Or potential energy (of molecules) increases and thekinetic

energy remains constant

temperature remains constant

B1

B1

(b) (i) P t= m c 48 x 720 = 0.98 x cx (54 - 18)

+0.027 x 850 x (38-18)

c= 970 (J kg

-1

K

-1

)

C1C1

C1

A1

Note: mark is for correand heat gained

Note: mark is for addingained by

insulation into thNote: answer to 3 sf = 9Calculation of c= 980 igscores 2 marks

(ii) Without the insulation there will be moreheat lost to thesurroundings / air (AW)

final temperature will be lowerORa lowertemperature riseORmore energy will be required to give the same

temperature rise / final temperature

AND hence c is higherthan that calculated in (i)

M1

A1

Not:lost to wires / data

Total 10


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142403 Mark Scheme Unit g484 the Newtonian World January