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January 2006

GCE

Edexcel GCE

Physics

Units PHY5 & PHY6

Supplement to UA017392

Mark Scheme

EdexcelG

CE

Physics(8

540/9540)

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Contents

Notes on the Mark schemes 1

Units PHY5/01 Mark scheme 2

Unit PHY5/02 (Practical test) Mark scheme 7

Unit PHY6 Mark scheme 15

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1

Notes on the Mark Schemes

1. Alternative responses: There was often more than one correct response to a particularquestion and these published mark schemes do not give all possible alternatives. Theygenerally show only the schemes for the most common responses given by candidates. Theyare not model answers but indicate what the Examiners accepted in this examination.

2. Error carried forward: In general, an error made in an early part of a question is penalisedthere but not subsequently, i.e. candidates are penalised once only, and can gain credit inlater parts of a question by correct reasoning from an earlier incorrect answer.

3. Quantity algebra: The working for calculations is presented using quantity algebra in themark schemes for Units PHY1, PHY2, PHY3 (Topics), PHY4, PHY5/01, and PHY6 butcandidates are not required to do this in their answers.

4. Significant figures: Use of an inappropriate number of significant figures in the theorypapers will normally be penalised only in show that questions where too few significantfigures has resulted in the candidate not demonstrating the validity of the given answer.Use of an inappropriate number of significant figures will normally be penalised in the

practical tests. In general candidates should nevertheless be guided by the numbers ofsignificant figures in the data provided in the question.

5. Unit penalties: A wrong or missing unit in the answer to a calculation will generally loseone mark unless otherwise indicated.

6. Quality of written communication: Each theory paper will usually have 1 or 2 marks for thequality of written communication. The mark will sometimes be a separate mark andsometimes be an option in a list of marking points.

Within the schemes:

/ indicates alternative marking point( ) brackets indicate words not essential to the answer[ ] brackets indicate additional guidance for markers

The following standard abbreviations are used:a.e. arithmetic error (1 mark)e.c.f. error carried forward (allow mark(s))

s.f. significant figures (1 mark only where specified)no u.e. no unit error

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2

6735 Unit Test PHY5

1. Calculation of capacitance

Either

Substitution of 1.08 10-4 J and 6 V into W =21 CTV

2 /21 QTV

Value of CT = 6 F / QT = 36 C

Use ofT

1

C=

C

2i.e.

F6

1

=

C

2/ C =

V

Qi.e.

V3

C36

Answer = 12 F

OR

If solution is obtained from considering single capacitor

Appreciation that energy is shared between capacitors i.e. W= 5.4 10-5

FThat voltage is distributed equally ie V= 3 V

Substitution into W =2

1CV2 i.e. 5.4 10-5 J =

2

1C(3 V)2

Answer = 12 F

9999

9999 4

Calculate the charge stored

CT = 6 F / Q3 = 3 F 6 V = 18 C

QT = CTV= 6 F 6V = 36 C / QT = 2 18 C = 36 C

99 2

Confirm energy storedis 108 J

2

1QV=

2

136 C 6V /

T

2

2C

Q=

( )F62

C362

[Use of2

1CV2 scores no marks]

9 1

7

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3

2. Changes in magnitude and direction of resultant force

Direction initially towards the Earth

Resultant magnitude decreasing to (zero)/ gravitational force of

Earth decreases that of moon increases

Resultant magnitude now increases

Direction towards the MoonThere is a point, nearer to the Moon, where the resultant force is

zero / fields or forces are equal and opposite

99

99

9 max 4

Gravitational force between masses

2

21

r

mGm[must use symbols given and G]

9 1

Relative size of gravitational forces

Values compared e.g.2

r

mGME 2

m

r

mGM=

M

E

M

M

[Allow one mark for ecf in line above]

Use of the masses to show the factorkg1035.7

kg1098.522

24

= 81.3

9

9

2

Distance to moon

Correct use of F =2

E

r

mGMi.e.

2.82 10-3N =2

2411

kg1kg1098.51067.6r

r= 3.76 108 m

9

9 2

9

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4

3. Directions and names of forces

Upward arrow labelled electrostatic/ force due to field/electric

force/ force of attraction

Downward arrow labelled weight / mg / gravitational force

[Not gravity]

9

9

2

Charge on the drop

Use ofd

Vi.e.

m)10(5.2

V5003

/ 2 (10

5) V m

-1

Use ofmg=d

VQ i.e.1.96(10

-14) kg 9.81 N kg

-1= 2 (10

5)Vm

-1Q

Q = 9.61 10-19

C

9

9

9

What happens to the drop ?

Drop will accelerate upwards

E/dV is increased

Hence mg or weight

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5

4. Why the reading increases

QOWC

A current in a magnetic field produces a force / produces a magneticfield.

This force acts upwards on the wire.

Its newton 3rd

law pair acts downwards on the magnet.

99

99

4

Magnetic flux density

Use ofmgi.e. 0.45 (10-3) kg 9.81 N kg-1

Use mg=BIL / 0.45(10-3

) kg 9.81 N kg-1

=B 1.5A6 (10-2

) m

Answer: 4.9 10-2 T

999

3

Procedure

Close the switch and note the values on the (ammeter) and balanceVary the voltage / adjust the variable resistor to obtain more values

GraphLabels and line must match

99

9

2

1

10

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6

5. Magnitude of vertical component

Vertical component = 4.0 10-5

T 9 1

Area drawn out

18 250 0.8 9 1

Flux change (ecf)

Use of vertical component i.e. 4 10-5 T

0.14(4) Wb / T m2

(s-1

) ( 4 10-5

T 3600 m2

)

99 2

Emf induced (ecf)

0.18 V i.e.s8.0

Wb144.0

9 1

Why no charge flow

(Wings are) not part of a closed circuit9

1

6

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7

6735/2A Practical Test PHY5

Question A

(a) All values sensible and with units 9

Repeats and averaged values forw and t 9

w and tto 0.01 mm and lto 1 mm or better 9 3

Zero error checked on balance or micrometer 9

Any

Careful not to stretch/squash band 2 9

Repeats taken at different places 9 2

Correct calculation of density (with m correct) 9

2/3 significant figures + unit 9

10% of Supervisors value 9 3

(b) (i) nT 20 (sensible value) 9

0.07 s of Supervisor [not if nTto nearest second] 9 2

(ii) Use of vertical rule [vertical stated or set square shown] 9

Difference method or measure at each end 9

Sensible value with unit 9 3

(iii) Sensible value and 2t or repeat for both t1 and t2 9

1.23 1.43 99

[1.13 1.53 only 1 mark] 3

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8

Sample results

(a) t: 2.84, 2.80, 2.82, t= 2.82 mm

w: 0.93, 0.92, 0.89, 0.92, w = 0.91(5) mm

l= 2 149 = 298 mm

m = 0.97 g

Zero error checked on balance or micrometer

Careful not to stretch/squash band

Repeats taken at different places

=v

m =0915.0282.08.29

97.0

= 1.26 g cm-3

(b) (i) 10 TA/s: 13.90, 13.98

TA = 1.39 s

(ii)

Use of vertical rule [vertical stated or set square shown]

Difference method or measure at each end

4t1 = 95.00 s t1 = 23.8 s

(iii) 4t2 = 70.60 s t2 = 17.7 s

2

1

t

t

= 7.17

8.23

= 1.34

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9

Question B

(a) All three voltages recorded to 3 significant figures or better +unit

V1V2

9

9 2

(b) Current is the same 9

Hence correct argument [dependent mark] 9

Correct calculation using mean as denominator 9 3

(c) All V values to 3 significant figures 9

1 /R correct and 2 or 3 decimal places and 1/ Vcorrect and 2 or

3 decimal places

9

All 7 values sensible 9

6 good values ( 2 mm from examiners best line) 999

[5 good values 2 marks; 4 good values 1 mark] 6

(d) Plots 9

Line 9 2

(e) Large triangle [base 8 cm] 9

Correct calculation [ignore unit] 9

Correctly calculation with 2/3 significant figures + unit 9 3

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10

Sample results

(a) V0 = 1.570 V

V1 = 0.787 V

V2 = 0.782 V

(b) Current is same inR1 andR2, so V/R is the same.

Percentage difference =

2

782.0787.0

782.0-787.0

+ 100 %

= 0.6%

(c) R / k V/ V (1 /R) / k1 (1 / V) /V11.0 0.231 1.00 4.33

1.2 0.259 0.83 3.86

1.5 0.298 0.67 3.36

2.2 0.370 0.45 2.70

2.5 0.392 0.40 2.55

2.7 0.408 0.37 2.45

3.7 0.463 0.27 2.16

(d) See graph

(e) s =00.000.135.135.4

= 3.00 k V1

X= 3.00 k V1 1.570 V

= 4.7 k

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11

(d)

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12

Question C

(a) (i) Thermistor in oil/water bath 9

Thermometer 9

Method for heating 9

Method for resistance 9 4

(ii) Correct re-arrangement intoy = mx + c 9 1

(b) ln values correct to 2 decimal places and with unit at head ofcolumn

9

GraphScale: page in each direction avoiding 3s etc 9

Axes: labelled with units 9

Plots: 1mm precision 9

Line: Straight and smooth curve 9 5

(c) (i) [20 C 55/65 C] sensible range with unit 9

Reason 9

(ii) Correct intercept 9

R0 + unit 9

Linearity assumed 9

(iii) Idea of using ice to get temperature of 0 C 9 6

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13

Sample results

(a) (i)

(ii) R =R0 ek

lnR = k+ lnR0 y = mx + c

(b) / C R / ln (R / )20 706 6.56

30 491 6.20

40 350 5.86

50 245 5.50

60 174 5.16

70 134 4.90

80 105 4.65

90 87 4.47

100 74 4.30

See graph

(c) (i) Between 20 C and 60 C where graph is linear.

(ii) ln (R0 / ) = 7.24

R0 = 1.39 k

Assuming graph remains linear between 20 C and 0 C

(iii) Pack ice round thermistor and record resistance when

temperature is 0 C.

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14

(b)

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6736 Unit Test PHY 6

Mark Scheme

1.(a) (i)

(ii)

(iii)

Superfluid: no resistance to flow / no viscosity / dissipate no energy as it

flows

Superconductor: no electrical resistance / zero resistivity / dissipate no

energy when carrying a current

Common feature: two of

both quantum phenomenon / involve wavelike nature of matter

both need very low temperatures

both dissipate no energy [notif given two marks in (i)]

Condense (paragraph 1): vapour / gas to liquid

condense (paragraph 4): varying / exchanging E

to commonE

9

9

99

4

9

9

2

(b) (i) p = mandE= m2

algebra to eliminate

p = (2mE)

9

9

2

(ii) E= (2.1 x 1023 J K

1)(1.8 K) / 3.78 1023 J

Substitute m = 4 1.66 1027 kg

p = 7.09 / 7.1 1025 N s / kg m s1

Use of = h p with h = 6.63 1034 J s / 6.6 1034 J s

= 9.3 / 9.4 1010

m (ie 1 nm)

Comment: 1 nm related to size of atom

9

9

9

9

9

9

max 5

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16

(c) (i) Two roughly sinusoidal waves of same in phase

Two roughly sinusoidal waves of different / broken

99

2

(ii) Quantum physics the wavelike nature of particles [c.e.p.]

Other phenomenon: the particle-like nature of photons

/ photoelectric effect / wavelike electrons in atoms

99

2

(iii) Experiment: electron gun / anodecathode

graphite / crystal / charged wire

mention of diffraction / superposition / interference

fluorescent screen

vacuum

99999

max 4

(d) (i) Graph: linear > 4 K

proportional > 4 K

nearly vertical at 4 K / 4.2 K

zero from 0 K to 4 K/ 4.2 K

9999

4

(ii) 195 oC = 78 K

Use of ratio 5 K 78 K OR proportional to Tused

5 K= 3.7 10

9 m

999

3

(e) Reflection of magnet shape

Reflection of poles

Like poles repel

999

3

31

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17

2.(a) (i) Circuit: cell /cells / d.c. supply

switch / two-way switch

A and R /16.5 k in series

999

3

(ii) Method involves area under graph

Count squares / draw triangle (i.e. correct use of method)

2.4 3.2

2.55 2.95

9999

4

(b) I= Q RC(allow minus)

SubstIand Q at given tOR use Q = Qo e at t=RCOR use half-life =

6.0 s OR lnQ = lnQ0 t/RC

RC= (8.5 0.5) s

R = above 470 106

F (about 18 k)

Subtract 16.5 k

99

999

5

(c) (i) Exponential / logarithmic 9(ii) Decay constant / 1n 2 half-life 9

2

(d) Experiment: (group) coin tossing / dice rolling / emptying burette

remove all heads / sixes each time / Vat fixed tintervals

Improve: increase number of coins / dice

999

3

17

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18

3.(a) (i) Use ofP=IVIin one is 0.5 A

So total current is 2 / 2.0 A

Assumption: no otherR in circuit / battery zero r/ metal frame zeroR

999

3

(ii) Use ofP= V2 R / P=IVandR = VI

one is 3 / 3.0

Two combined in parallel 1.5

c.f. car battery rvery small / 0.02 / a few m / zero

9

99

3

(b) (i) QOWC (accept bullet points / phrases)

Current in P makes P electromagnet / produces magnetic field / produces

magnetic flux (in core)

Switch C opened / current switched off leads to change in B /

(through S)

e.m.f. / (induced) voltage

AsNs much greater thanNp / tvery small inN t/ step up voltagetransformer

Detail: large electric field in gap / air ionised in gap / c.e.p.

99

9

99

9

max 5

(ii) Use ofE= V/d

V= (3.2 10

6

V m

1

)(0.85 10

3

m) [ignore 10

n

]

= 2700 V / 2.7 kV

999

3

14

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19

4.(a) (i) Average speed 1.0 m s1 1.3 m s1

Average speed 1.1 m s1 and 1.2 m s1

Use ofav =s/t s = (their value av)(15 s)

9993

(ii) Slows down / decelerates (from 1.55 m s1 to 0.60 m s1)

As no / little thrust

Water drag / resistance

So resultant force / push of water is backwards

9999

max 3

(iii) Attempt at tangent at or near 0.90 s

Acceleration calculated using triangle with t 0.4 s

a = 2.5 2.9 m s2

Resultant force = (their value)(65 kg) with unit N

9999

4

(iv) Graph: passes through zero at 0.3 s and0.8 s

is zero at start and end of stroke

two +ve bits and one ve bit

999

3

(b) (i) Distance / circumference = 2y0

c =f orT= 1/f

= 2f0c (no mark)

99

2

(ii) Gain of g.p.e. / p.e. = mg2y0

Loss of k.e. = m(c + )2 m(c )2

Equated to yield 2mgy0= 2mc

c =(g/2) (no mark)

999

max 3

18