Mark Scheme January 2007

GCE

GCE Physics (8540/9540) International

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January 2007

All the material in this publication is copyright Edexcel Ltd 2006

i

Contents

Page Mark Scheme notes 1 Unit PHY1 Mark Scheme 3 Unit PHY2 Mark Scheme 11 Unit PHY3/01 (Topics) Mark Scheme 19 Unit PHY3/02 (Practical Test) Mark Scheme 27 Unit PHY4 Mark Scheme 35 Unit PHY5/01 Mark Scheme 43 Unit PHY5/02 (Practical Test) Mark Scheme 51 Unit PHY6 Mark Scheme 59

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1

Mark scheme notes Underlying principle The mark scheme will clearly indicate the concept that is being rewarded, backed up by examples. It is not a set of model answers. For example:

(iii) Horizontal force of hinge on table top 66.3 (N) or 66 (N) and correct indication of direction [no ue] [Some examples of direction: acting from right (to left) / to the left / West / opposite direction to horizontal. May show direction by arrow. Do not accept a minus sign in front of number as direction.]

9 1

This has a clear statement of the principle for awarding the mark, supported by some examples illustrating acceptable boundaries. 1. Mark scheme format

1.1 You will not see wtte (words to that effect). Alternative correct wording should be credited in every answer unless the ms has specified specific words that must be present. Such words will be indicated by underlining e.g. resonance

1.2 Bold lower case will be used for emphasis. 1.3 Round brackets ( ) indicate words that are not essential e.g. (hence) distance is

increased. 1.4 Square brackets [ ] indicate advice to examiners or examples e.g. [Do not accept

gravity] [ecf]. 2. Unit error penalties

2.1 A separate mark is not usually given for a unit but a missing or incorrect unit will normally cause the final calculation mark to be lost.

2.2 Incorrect use of case e.g. Watt or w will not be penalised. 2.3 There will be no unit penalty applied in show that questions or in any other question

where the units to be used have been given. 2.4 The same missing or incorrect unit will not be penalised more than once within one

question but may be penalised again in another question. 2.5 Occasionally, it may be decided not to penalise a missing or incorrect unit e.g. the

candidate may be calculating the gradient of a graph, resulting in a unit that is not one that should be known and is complex.

2.6 The mark scheme will indicate if no unit error penalty is to be applied by means of [no ue].

3. Significant figures

3.1 Use of an inappropriate number of significant figures in the theory papers will normally only be penalised in show that questions where use of too few significant figures has resulted in the candidate not demonstrating the validity of the given answer.

3.2 Use of an inappropriate number of significant figures will normally be penalised in the practical examinations or coursework.

3.3 Using g = 10 m s2 will not be penalised.

2

4. Calculations 4.1 Bald (i.e. no working shown) correct answers score full marks unless in a show that

question. 4.2 If a show that question is worth 2 marks then both marks will be available for a

reverse working; if it is worth 3 marks then only 2 will be available. 4.3 use of the formula means that the candidate demonstrates substitution of physically

correct values, although there may be conversion errors e.g. power of 10 error. 4.4 recall of the correct formula will be awarded when the formula is seen or implied by

substitution. 4.5 The mark scheme will show a correctly worked answer for illustration only. 4.6 Example of mark scheme for a calculation:

Show that calculation of weight Use of L W H Substitution into density equation with a volume and density Correct answer [49.4 (N)] to at least 3 sig fig. [No ue] [Allow 50.4(N) for answer if 10 N/kg used for g.] [If 5040 g rounded to 5000 g or 5 kg, do not give 3rd mark; if conversion to kg is omitted and then answer fudged, do not give 3rd mark] [Bald answer scores 0, reverse calculation 2/3] Example of answer:

80 cm 50 cm 1.8 cm = 7200 cm3 7200 cm3 0.70 g cm-3 = 5040 g 5040 10-3 kg 9.81 N/kg = 49.4 N

9

9

9

3

5. Quality of Written Communication

5.1 Indicated by QoWC in mark scheme, placed as first mark. 5.2 Usually it is part of a max mark. 5.3 In SHAP marks for this are allocated in coursework only but this does not negate the

need for candidates to express themselves clearly, using appropriate physics terms. Likewise in the Edexcel A papers.

6. Graphs

6.1 A mark given for axes requires both axes to be labelled with quantities and units, and drawn the correct way round.

6.2 Sometimes a separate mark will be given for units or for each axis if the units are complex. This will be indicated on the mark scheme.

6.3 A mark given for choosing a scale requires that the chosen scale allows all points to be plotted, spreads plotted points over more than half of each axis and is not an awkward scale e.g. multiples of 3, 7 etc.

6.4 Points should be plotted to within 1 mm. Check the two points furthest from the best line. If both OK award mark. If either is 2 mm out do not award mark. If both are 1 mm out do not award mark. If either is 1 mm out then check another two and award mark if both of these

OK, otherwise no mark. 6.5 For a line mark there must be a thin continuous line which is the best-fit line for the

candidates results.

3

6731 Unit Test PHY1

1. How A and B change Force B For ticking no change in all 4 boxes Force A 4 ticks right 999 3 ticks right 99 2 ticks right 9

increases no change decreases 9 9 9 9

9 9 9 9

4

4

2 a Path of coin

Curved line that must begin to fall towards the ground immediately

9

1bi Show that..

Selects s = (ut +)21 at2 or selects two relevant equations

Substitution of physically correct values into equation or both equations. Answer [0.37 s - 0.38 s] [Allow use of g = 10 m s-2. Must give answer to at least 2 sig. fig., bald answer scores 0. No ue.]

eg 0.7 m = 21 (9.81 m s-2 )t2

t = 0.38 s

9 9 9

3

bii Horizontal distance [ecf their value of t]

Use of v = td with correct value of time. [s =

2uv+ t is sometimes

used. In this case v and u must be given as 1.5 m s-1 and t must be correct. Also s = ut + 0.5at2 OK if a is set = 0.] Answer [0.55 m 0.60 m] eg d = 1.5 (m s-1) 0.38 (s) = 0.57 m

9 9

2

c A coin of greater mass? QWOC It will follow the same path [accept similar path, do not accept same distance] All objects have the same acceleration of free fall / gravity or acceleration of free fall / gravity is independent of mass / it will take the same time to fall (to the floor) Horizontal motion / velocity is unaffected by any force or (gravitational) force (acting on coin) has no horizontal component or horizontal motion/velocity is the same/constant.

9 9 9

9

4

10

5

3. a Meaning of 0.8 s

Reaction time (of cyclist and car driver) [Accept descriptions of reaction time eg time it takes both to take in that the lights have changed to green. Accept response time]

9

1

bi Same speed time

Answer [6.8 s - 6.9 s] [Accept any value in the range]

9

1bii How much further ahead?

Either For measuring area under car graph at 6.8 s

eg 2

sm9s6 1 = 27 m [27.5 m if 6.9 s used] For measuring area under cyclist graph at 6.8 s

eg 2

sm9s2 1 + 4 s 9 m s-1 = 45 m [45.9 m if 6.9 s used] [For candidates who read the velocity 9 m s1 as 8.5 m s1 but otherwise do their calculation(s) correctly give 2/3] [Allow one mark to candidates who attempt to measure an appropriate area] Answer [( 45 m 27 m =) 18 m] Or For recognising the area enclosed by cyclist and car graphs as the difference in distance travelled Using values from the graph to determine this area Answer [(45 m 27 m =) 18 m]

eg distance = 21 (6.8 2.8) s 9 m s-1

= 18 m

9

9

9 9 9 9

3

c Relationship between average velocities

They are the same

9

1 6

6

4 a Calculate the weight

Use of L x W x H Substitution into density equation with a volume and density Answer [ 466 N or 475 N] [466(.17) N or 475 N if 10 N kg-1 is used. At least 3 sig fig required, no ue. It must be clear that either 9.8(1) N kg-1 or 10 N kg-1 has been used. Bald answer scores 0] eg 0.6 m 0.6 m 0.04 m = 0.0144 m3 0.0144 m3 3300 kg m-3 = 47.52 kg 47.52 kg 9.81 kg m-3 = 466 N

9 9 9

3

bi Position of centre of gravity Position should be seen to have been found by drawing eg intersection of diagonals between opposite corners

9

1bii Force applied

Use of principle of moments Answer [230 N 235 N] eg F 0.6 m = 470 N 0.3 m [Allow use of d and 2d or equivalent] F = 235 N

9 9

2

c Why it becomes easier (Because perpendicular) distance from weight (direction) to pivot reduces/(component of) weight perpendicular to the slab reduces The moment of the weight (about AB) reduces (as slab is raised) or clockwise moment reduces (Since perpendicular) distance from (applied) force (direction) to pivot remains constant (Hence applied) force (can be) reduced (to overcome smaller weight moment) or anticlockwise / (applied) force moment is reduced [Candidates should refer to weight or force to obtain each of these marks. References to mass and centre of gravity alone are insufficient.]

Or Initially the height of the centre of gravity (of the slab) increases more or initially the gpe of the slab increases more For a given / the same distance/angle moved by the slab This means more work is done by the (applied) force (lifting the weight of the slab) (Hence applied) force reduces (as the slab is raised to the vertical)

9

9 9 9

Max 3

9 9 9 9

Max 3

9

7

5 a Momentum at impact p = mv seen or used Answer [11 kg m s-1] eg momentum = 0.42 kg 27 m s-1 = 11.34 kg m s-1

9 9

2

b Momentum at release Minus 8.4 kg m s-1

9 9

2ci Average force(ecf momenta values)

Use of F = tp ie for using a momentum value divided by

0.22 Adding momentum values Answer [88.0 N - 89.8 N]

F = s 22.0

s m kg3.11s m kg4.8 11 F = () 89.5 N Or Use of F = ma Adding velocities to calculate acceleration Answer [88.0 N 89.8 N]

Eg acceleration = s22.0

sm27sm20 -1-1 ( = 213.6 m s -2) Force = 0.42 kg 213.6 m s-2 = ()89.7(2) N

9 9 9 9 9 9

3cii Direction of force on diagram

Right to left [Accept arrow drawn anywhere on the diagram. Label not required]

9

1

d Difference and similarity Difference: opposite direction / acts on different object Similarity: same type of force / same size / acts along same line / act for same time / same size impulse [Magnitude and size on their own is sufficient. They are equal is OK. Accept; they are both contact forces; they are both electrostatic forces]

9 9

2

10

8

6 a EK of helium nucleus

Use of EK = 21 mv2

Answer [ 3.1 10-15 J. No ue. Min 2 sig fig required]

eg EK = 21 6.65 x 10-27 kg ( 9.65 105 m s-1)2

= 3.096 10-15 J

9 9

2

bi Loss of EK of proton [ecf their value for EK of helium nucleus] 3 10-15 J or 3.1 10-15 J

9

1bii Speed of proton after collision

[ecf their value for loss of EK of proton, but not if they have given it as zero] Calculation of initial Ek of proton Subtraction of 3.1 10-15 J [= 1.7 10-15 J] Answer [(1.40 1.50) 106 m s-1]

eg EK = 21 1.67 10-27 kg x (2.4 106 m s-1)2 (= 4.8 10-15 J)

EK after collision = 4.8 10-15 J - 3.1 10-15 J ( =1.7 10-15 J)

v = (gk1067.15.0

J107.127

15

)0.5 = 1.43 106 m s-1

Or Use of the principle of conservation of momentum. Correct expression for the total momentum after the collision Answer [ (1.40 1.50) 106 m s-1] Eg 1.67 10-27 kg x 2.4 106 m s-1 = 6.65 10-27 kg (9.65 105 m s-1) + 1.67 10-27 kg V V = 1.44 106 m s-1

[For both these solutions allow the second marking point to candidates who incorrectly write: the mass of the proton as 1.6 1027 kg or 1.7 1027 kg, or the mass of the helium as 6.6 1027 kg or 6.7 1027 kg or the velocity as 9.6 105 m s1 or 9.7 105 m s1]

9 9 9

9 9 9

3

c Other factor conserved Momentum / mass / charge / total energy

9

1

7

9

7 ai aii

Deductions The atom is mainly empty space [The atom must be referred to. The words empty and space must be qualified eg there is a large amount of space in the atom is not sufficient] Within the atom there is an area / the nucleus which is positive / charged or more massive than the alpha particle [If they choose to describe only the mass it must be a comparison ie the nucleus is (much) more massive than the alpha. The atom has a dense centre, the nucleus has a large mass are both insufficient.]

9 9

2 b Explain

(Deflection could have been) repulsion from positive nucleus (Deflection could have been) attraction towards negative nucleus [The words repulsion and attraction can be described eg deflected away from positive nucleus, is deflected towards a negative nucleus] [Diagrams showing the path of an alpha deflected by both a negatively and a positively charged nucleus would get both marks]

9 9

2

c Value of n (4 6) [Allow minus values]

9

1

5

10

8a Meaning of random

Impossible to predict which atom/nucleus (in a given sample) will decay (at any given moment)/ unable to predict when a given atom will decay [Mention of atom(s), or nucleus, or nuclei is essential because the word random is to be described in context. Do not accept for atom or nucleus; substance; material; particle; molecule; sample.]

9

1

b Nuclear equation 24195 Am 23793 Np + 42He

42He or

42

24195 Am /

23793 Np / both proton numbers correct / both mass numbers

correct Entirely correct equation

9 9 9

3

c Absorbtion experiment Diagram [must include the source, detector and an indication of where the absorber is placed (maybe written in their account rather than on the diagram) none of these need to be labelled ] (Record) background count Source detector distance must be close / less than or equal to 2 cm Insert eg paper between source and detector = no change in count rate or increase (by a small amount) the separation of the detector and source = no change in count rate (therefore no present) Insert aluminium/brass (a few mm thick) or lead ( 2mm thick) / concrete Count reduced to background (therefore no gamma present) [Do not give this mark if only paper or card or plastic is used as the absorber. Accept 0 in place of reduced to background if candidate has deducted background from their measurements.]

9 9 9 9 9

9

Max 5

9

11

6732 Unit Test PHY2

1 Word equations [Must be in words for credit.] Force / Area 9 Resistance x 9 Cross sectional area /length 9 absolute temperature x number of mole(s) x (molar) gas constant Product of any two 9 Multiplied by the third 9 5 Accept temperature in Kelvin or temperature/K Accept mass/molar mass Do not accept just moles

12

2 Tungsten filament bulb a Resistance Use of P=V2/R or P=VI with V=IR 9 answer 960 9 2 Example of answer R = (240 V 240 V) 60 W R = 960 b Drift speed rearrangement of I = nAvQ 9 Use of Q = 1.6 x 10-19 (C ) 9 answer 0.15/0.148 m s-1 9 3 Example of answer v = 0.25 A ( 3.4 10 28 m-3 1.6 10 19 C 3.1 10 -10 m2) c Explanation Qowc 9 Any THREE Resistance due to collisions between electrons &

ions/atoms/particles

(as T increases) ions/atoms/particles have more energy (as T increases) ions/atoms/particles vibrate through larger

amplitude /vibrate faster OR amplitude if lattice vibration increases.

more chance/increased frequency of collision/interaction OR impedes the flow of electrons

999 4 9

13

3 Emf and Internal resistance a Derivation E = I (R + r) OR E = IR + Ir 9 1 bi Correct working (allow even if evidence of working backwards) 9 1 Example of answer E/I = R + r Rearranging R = E/I r ii Emf Attempt to use gradient 9 answer 1.5 V (bald answer 1.5 V scores 0/2) 9 2 iii Power From graph find value of 1/I when R = 5 9 Use of P = I2R 9 answer 0.31 (W) 9 3

Example of answer

1/I = 4 A1 I = 0.25 A

P = 0.25 A 0.25 A 5 = 0.3125 W

c Graph Intercept at 2 (ohms) 9 Graph steeper than original 9 Gradient is 3.0 V i.e. line passes through [10, 27-29] [no ecf] 9 3 10

14

4 Potential divider a First circuit Middle terminal M Outer terminals L and K (any order) 9 1 bi P.d across lamp. External resistance in circuit is 25 or (20+5) ohms 9 See ratio of resistances (denominator larger) x 6.0 V

OR current = 6/25 A 9

answer 4.8 V 9 3 ii Assumption The resistance of the ammeter is zero/negligible. 9 1 c Second circuit See 2 resistors in parallel with supply 9 Supply across ends of variable resistor (10 ) 9 Fixed resistor across one end and slider (consequent mark) 9 3 8

15

5 Latent heat ai Apparatus and circuit. Supply connected to immersion heater/heating coil in

water or kettle

Clock / balance under beaker/ beaker under apparatus Power rating of immersion heater given

or A & V correctly connected or joulemeter connected to supply

Start clock when boiling is established Read Joulemeter when boiling starts

After measured time obtain change of mass (condensed or evaporated) OR 2nd Joulemeter reading

VIt or Pt or reading from Joulemeter = mL 9999

4 LH fusion of ice max 3 Experiments recording temp rise max 2 ii Precaution Any sensible suggestion for their arrangement, for example Water covers heating coil Tall beaker to avoid spillage Use of lagging Partial lid to avoid spillage 9 1 bi Energy Conversion of g to kg 9 168000 / 1.68 105 (J) / 0.17 (MJ) 9 2 Example of answer E = 0.4 kg 4200 J kg 1K1 100 K = 168000 J ii Energy 920000 / 9.2 105 J / 0.92 MJ 9 1 Example of answer Energy = 0.4 kg 2.3 106 J kg1 iii Graph Straight line graph from (0,0) to (100, 1.7 or 2.0) 9 Vertical line at 100 C of length 9 / their value 9 2 10

16

6 Heat engine a Symbols T1 Temperature of the hot source/reservoir/body 9 T2 Temperature of the cold sink/reservoir/body 9 (T1 initial temperature AND T2 final temperature 1 mark only) W Work done by engine OR mechanical work done OR useful

work done 9

Q2 Wasted energy OR remaining energy that flows to sink 9 4 ecf wording from T2 bi Temperature calculation Uses Efficiency = (T1 - T2 )/T1 (K or 0C) 9 Conversion to Kelvin ie see 500 K 9 answer 380 K (107 C ) 9 3 Example of answer E = 1 T2/ T1 T2/T1 = 1 0.24 T2 = 0.76 500 K = 380 K bii Efficiency By increasing T1 /temperature of the hot steam Or by decreasing T2 / temperature of the cold sink Or by increasing the temperature difference. 9 1 8

17

7 Gas laws a Boyles law Uses pV = constant 9 See V2 = 15 or (20 5) cm3 9 Pressure = 267/270 x 103 Pa /Nm-2 9 3 Example of answer p1V1 = p2V2 p2 = 200 103 Pa 20 cm3 15 cm3 p2 = 266667 Pa b Force Uses F = pA 9 15.8/16.0 N 9 2 Example of answer F = 200 103 Pa 7.9 105 m2 F = 15.8 N ci Pressure law Uses p/T = constant Kelvin or Celsius. 9 At least one conversion to Kelvin (295K or 308K) 9 209 x 103 Nm-2/Pa 9 3 Example of answer p1/ T1 = p2 / T2 p2 = 200 103 Pa 308 K 295 K p2 = 208813 Pa ii Graph Curve reasonably similar to one given 9 Curve above first. (no ecf for their p2 less than p1) 9 2 10

18

19

6733 Unit Test PHY3 Topic A - Astrophysics

(a) Revision notes errors

Red giant cepheid (variable) White dwarf red (super)giant / supernova / core remnant > 2.5 M~ Red giant main sequence (star)

99 99 99 6

(b) (i)

Wiens law Wavelength of peak [or maximum] intensity [allow brightness, but not power / energy / L] Absolute or Kelvin (surface) temperature (of star)

9 9 2

(ii) (iii)

Spectrum Microwave / infra-red [accept i.r.] max = 1.05 (mm) Substitution in T = 2.90 10-3 m K their max with 103 [2.90 10-3 m K 1.05 10-3 m, with mm m conversion required] T = 2.76 [accept range 2.6 2.9] K

9 9 9 9

1

3 (c) (i)

Supernova minimum mass 8 M~ [ accept 1.6 10

31 kg]

9 1

(ii) (iii)

Energy from Sun 1 1010 365() 24 60 60 / 3(.15) 1017 Use of E = P t

1(.2) 1044 J [Beware E = tP 1.24 109 J]

E = P t = 3.9 1026 W 1 1010 365() 24 60 60 s = 1.2 1044 J 1046 (1.2 x) 1044 [ecf for any E] 80 100 [ecf] [accept 83:1 or 1:0.012] [inverted answer scores zero, unless values identified for 1/2]

9 9 9 9 9

3

2

20

(iv)

Supernova future neutron star / pulsar black hole n.s. if > 1.4 M~ OR b.h. if > 2.5 M~

9 9 9 3

(d) (i)

Neutron star density = m V and 4/3 r3 4.2 [or 4.3] 1017 (kg m-3) = m V = 3m 4 r3 = 3 6.0 1024 kg (4 (150 m)3) = 4.2 1017 kg m-3

9 9 2

(ii)

Neutron formation Quality of written communication Main sequence: fusion (reaction) / (ms) p n / beta plus decay [post ms] p + e n [post ms] due to gravitational collapse / implosion

9 9 9 9 4

(e) (i)

Intensity of Sun Use of I = L (4 D2) 597 OR 1380 (ignore 10n) 597 W m-2 AND 1380 W m-2 [accept W km-2 with appropriate values] I = L (4 D2) IM = 3.90 1028 W (4 (2.28 1011 m)2) = 597 W m-2 IE = 3.90 1028 W (4 (1.50 1011 m)2) = 1380 W m-2

9 9 9

3

597 1380 [ecf, accept (2.28 1.50)2] 43%

9 9 2

Total 32

21

Topic B - Solid Materials

(a) Revision notes errors Plastically elastically Rigid thermosets thermoplastics Fibre particle

99 99 99 6

(b)

(i)

Energy density units J m-3 [or energy equation reduced to correct base units] N m-2 [or stress shown to reduce to kg m-1 s-2] ll / no units for strain AND J = N m / correct mathematical reduction

9 9 9 3

(ii)

Energy density calculation Use of strain = ll Substitution in 1.0 106 (J m-3) E.D. = = ll = 5.2 108 Pa 8.0 10-3 m 2.0 m = 1.04 106 J m-3

9 9 9 3

(iii)

Energy density of mild steel Any attempt at area under graph 300 106 Pa 0.15 4.5 107 (J m-3) [accept range 4 5 with 107]

9 9 9 3

(c)

(i)

Property definitions Tough: absorbs energy (before breaking) by plastic deformation Strong: high(er) UTS / high(er) stress (before breaking)

9 9 9 3

22

(ii)

Force calculation Attempted use of = F / A [accept use of r instead of A for 1/3] Use of A = r2 (ignore 10n) 3.8 (or 4) 10-3 N = F / A = F / r2 F = r2 = 3 108 Pa (2.0 10-6 m)2 = 3.8 10-3 N

9 9 9 3

(iii)

Stiffest part of curve Initial slope indicated

9 1

(iv)

Young modulus calculation Any attempt at gradient / stress strain Correct pair of values for linear region above stress 0.25 / Extended gradient at start of curve 5.5 GPa [5.2 5.6 with GPa or GNm-2]

9 9 9 3

(d)

(i)

Hardening techniques Quality of written communication WH: repeated beating / hammering / bending [not working] QH: heat, then cool rapidly

9 9 9 9 4

(ii)

(iii)

Effect of work hardening Any two from: More brittle, harder, stiffer [or less flexible] Dislocations entangle [or description of this]

99 9

2

1

Total 32

23

Topic C - Nuclear and Particle Physics (a) Revision notes errors

Muon meson [accept pion or kaon] Strange top [all three keywords may be changed] Neutrons protons

99 99 99 6

(b)

(i)

Neutron star density = m V and 4/3 r3 4.2 1017 (kg m-3) = m V = 3m 4 r3 = 3 6.0 1024 kg (4 (150 m)3) = 4.2 1017 kg m-3

9 9 2

(ii)

Neutron number 6.0 1024 1.66 10-27 3.6 1051

9 9 2

(iii)

Neutron radius Attempted use of r = r0 A [or from first principles, ecf] 9.8 10-16 m / 1.0 10-15 m r0 = r / A = 150 / (3.6 1051) = 9.8 10-16 m

9 9

2

(iv)

Neutron formation 11 p +

01 e

10 n +

Correct symbols Correct values [ignore neutrino numbers]

9 9

2

(v)

Fundamental interaction explanation Quality of written communication weak change of quark flavour / u d / uud udd involves neutrino

9 9 9 9 4

24

(c)

(i)

Exchange particle gluon

9 1

(ii)

Delta particle charge (+)2 Conservation of charge and p = +1 / [or (+2) = (+1) + (+1)]

9 9 2

(iii)

(iv)

Quark compositions p = uud B: +1 = +1 + 0 hence baryon _ Cancelling of d d shown or explained uuu

9 9 9 9

1

3

(d)

(i)

Energy spectrum graph Correct shape: starts on y-axis [allow origin], peaks and asymmetric Reaches x-axis

9 9 2

(ii)

Additional particle explanation Spread of energies observed Conservation of energy expected Missing energy carried away by other particle

9 9 9 3

(iii)

Lepton name (anti) neutrino electron antineutrino [accept anti electron neutrino] _ [e only scores 1/2]

9 9 2

Total 32

25

Topic D - Medical Physics

(a) Revision notes errors Amplitude brightness / intensity of pixel Diagnosis therapy / destruction of tissue Large short /small

99 99 99 6

(b)

(i)

(ii)

Technetium use Metastable Any two of 6 hour half-life not too short for study nor to too long for added dose gamma (only) emitter minimises dose to patient / cell damage gamma emitter can be detected out of body / by gamma camera very long half-life daughter negligible harm to patient

9 99

1

2

(iii)

Different half-lives Quality of written communication

9

tb = time for activity to half due to biological processes / excretion te due to combined effects of tr and tb / time to half activity within body Reference to 1/te = 1/tr + 1/tb

9 9 9 4

(c) Gamma camera 1. Collimator Absorbs photons / gammas not at 900 to the camera 2. Photomultiplier tube Amplifies (number of) electrons 3. (NaI) Scintillator (crystal) produce light (flashes) / photons [ecf on all names if in incorrect box]

9 9 9 9 9 9 6

26

(d)

(i)

Transmitted percentage 70%

9 1

(ii)

Specific acoustic impedance calculation Correct substitution: 1.63 106 kg m-2 s-1 (1 + 0.30) (1 - 0.30) 5.58 / 5.6 [accept 4.76] (5.6 ) 106 kg m-2 s-1 [conditional mark, with ecf on use of 0.30 of 4.76]

9 9 9 3

(iii) Speed of sound calculation Substitution in c = Z [c = 1.63 106 kg m-2 s-1 1060 kg m-3] 1540 m s-1

9 9 2

(e)

(i)

MeV X-ray energies for therapy Absorption not dependent on proton number / independent of medium (Sufficient energy to) destroy / kill cells

9 9 2

(ii)

Multiple beam explanation Clear diagram with at least two beam paths, focused on labelled target Tumour / target cells always hit OR greater dose where beams cross Lower dose to healthy / surrounding tissue

9 9 9 3

(iii)

X-ray intensity Use of I = P 4r2 or I d2 = k [8.0 105 W m-2 4 (0.5 m)2 / P = 2.5 x 106 W / k = 2.0 105 W] 2.0 kW m-2

9 9 2

Total 32

27

6733/2A Practical Test PHY3

Question A

(a)(i) Circuit set up correctly without help 33 [If Supervisor corrects fault after readings taken 1] [All zero current given 0 marks here] 2 (ii)&(iii)

Units 3

All I to 0.1 mA or better 3 All V to 0.01 V or better and sensible 3 AC AB 3 CA CB 3 BC smallest current 3 RBA = 200 240 to 3 s.f. 3 7 (iv) Positive terminal of power supply connected to N and negative

terminal of power supply connected to M 33

[If other way round ie M and N then allow 1/2] Two paths for the current (N M and N L M)

or other resistors in parallel with R 3

so must have a resistance of less than R 3 max 2 All other cases must have resistance > R or all other paths contain

at least R + diode resistance or this path only contains R 3

[If MN chosen: Only one component between M and N or no

conducting diode 3]

Terminal L corresponds to C

Terminal M corresponds to B All correct Terminal N corresponds to A

3

[If positive terminal connected to A and negative connected to B,

then allow 3 for statement of greatest current or calculation of least resistance.]

5

28

(b)(i) Sensible values all to at least mm precision 3 All repeated 3 Correct calculation of density with unit 3 Value 0.30 to 0.80 g cm-3 and 2/3 s.f. 3 [No ecf] 4 (ii) 4 sensible readings to mm precision correctly averaged with unit 3 Pins in such a position that they counterbalance each other 3 Pins positioned to give correct average 3 3 (iii) Correct calculation, no unit, 2/3 s.f. 3 [Accept percentage] Value 0.10 of previous relative density 33 [ 0.20 of previous relative density 3] 3 24

29

Sample results

(a)(iii) X connected to Y connected to I / mA V / V A B 33.6 5.79 A C 16.8 5.89 B A 26.7 5.83 B C 8.9 5.94 C A 11.9 5.92 C B 23.2 5.85

(b)(i) l = 10.5, 10.6, 10.55 cm l = 10.55 cm w = 6.8, 6.75, 6.85 cm w = 6.8 cm t = 4.2, 4.25, 4.15 cm t = 4.2 cm V = 10.55 6.8 4.2 = 301.3 cm3 Density = 3cm301.3

g173 = 0.57 g cm-3

(ii) tB = 3.4, 4.2, 2.0, 1.4 cm

tB = 2.75 cm

(iii)

65.02.475.2B ==

tt

30

Question B

(a) Sensible h recorded to nearest mm or better 3 Correct calculation yielding sensible , 2/3 s.f. + unit 3 2 (b) Sensible t from 3 readings to at least 0.1 s + unit 33 [2 readings 3] Correct calculation 2 s.f.+ unit 3 [Allow ecf] 3 (c) Correct calculation 2 s.f. no unit and positive 3 [No ecf from incorrect or a] 1 (d) Larger h recorded to nearest mm or better + unit [ecf unit] 3 Correct calculation of 2 s.f. + unit [ecf unit] 3 Sensible t from 3 readings + unit [ecf unit]

[ 2 readings 3] [Sensible t: not whole seconds, not systematic error eg 0.0230, significantly less than t from part (b)]

33

Correct calculation of a and R 2 s.f., no unit and positive [ecf] 3 5 (e) Sensible value between 0.05 and 0.2 s 3 Correct calculation of percentage 3 2 (f) Correct calculation with average as denominator

(% symbol not required) 3

Doubled 3 Doubled again 3 Sensible comment based on percentage uncertainty calculation and

percentage difference 3

4

31

(g) Vary the height of the end of the runway / vary / use more blocks 3 Measure time (for trolley to travel given distance) 3 5 or more readings stated or seen 3 [Do not accept several/multiple. Accept many] Plot a against sin 3 Straight line with positive intercept on sin axis shown 3 Intercept on a axis = gR 3 R = Intercept / g 3 7 [Ignore negative sign in last 2 parts]

32

Sample results

(a) h = 24 mm l = 1175 mm tan =

117524 = 0.0204

= 1.17

(b) t = 3.15, 3.15, 3.26, 3.27 s t = 3.21 s a = 2

2tx = 221.3

8.02 = 0.156 m s2

(c) R = 0.0204

81.9156.0 = 4.5 10-3

(d) h = 43 mm, l = 1175 mm tan =

117543 = 0.0366

= 2.10 t = 2.27, 2.32, 2.33, 2.27 s t = 2.30 s a = 23.2

8.02 = 0.303 m s2

R = 0.0366

81.9303.0 = 5.7 103

(e) Range = 2.33 2.27 = 0.06 s or Reaction time = 0.2 s Percentage uncertainty =

30.22.0 100%

= 8.7%

33

(f)

Percentage difference = ( )( ) 33

105.47.5105.47.5

+ 100%

= 24% Percentage uncertainty in t2 = 17.4% But this uncertainty in both values of R, uncertainty in difference

= 34.8%.

difference can be explained by percentage uncertainty in t2.

34

35

6734 Unit Test PHY4

1.(a) Why transverse waves can be polarised but not longitudinal waves [Marks can be earned in diagram or text] Transverse waves have * perpendicular to direction of **

* = vibration/displacement/oscillation/motion of particles ** = travel/propagation/motion of wave/energy transfer/wave

9

In a transverse wave, * can be in different planes but polarisation restricts it to one plane

9

Longitudinal waves have * parallel to ** 9 3 [Dont accept motion for **

Diagrams to earn marks must be clearly labelled, but dont insist on a label looking along direction of travel in the usual diagrams to illustrate polarised and unpolarised waves]

(b)(i) Effect of Polaroid on intensity Intensity is reduced (OR halved) [not zero]

[Accept slightly reduced and greatly reduced] 9

Polaroid stops (OR absorbs) vibrations (OR waves OR light) in one plane/ Polaroid only lets through vibrations (OR waves OR light)in one plane/ Light has been polarised

9

2 (ii) Effect of rotating Polaroid No effect 9 [ignore incorrect reasons accompanying statements of effect] 1 6

36

2.(a) Calculation of efficiency Use of incident power = intensity x area 9 Use of efficiency = electrical power/incident power

[Or ratio of powers per unit area, or powers per cell] 9

Correct answer [0.18 or 18%] 9 3 e.g. Incident power = 1.4 kW m-2 20000 10 10-4 m2

= 28 kW Efficiency = 5 kW / 28 kW = 0.18

[Omission of 20000 loses marks 1 and 3]

(b)(i) Calculation of intensity Use of I = P / 4 r2 9 Correct answer [3.1 10-13 W m-2 OR 3.1 x 10-16 kW m-2

OR 3.1 x 10-7 W km-2 OR 3.1 x 10-10 kW km-2] 9

2 e.g. I = 5 103 W / (4(3.6 107m)2)

= 3.1 10-13 W m-2

[Failure to square r when substituting loses both marks] [Omission of 4 loses both marks] (ii) Calculation of intensity of focused beam Use of area = r2 9 Correct answer [6.4 10-9 W m-2 OR 6.4 x 10-12 kW m-2

OR 6.4 x 10-3 W km-2 OR 6.4 x 10-6 kW km-2] 9

2 e.g. I = 5 103 W / ((500 103 m)2)

= 6.4 10-9 W m-2

[Failure to square r when substituting loses both marks Use of diameter instead of radius loses both marks] [Inclusion of 4 or any other number loses both marks] [Missing or incorrect unit should be penalised only once in part (b)]

7

37

3.(a) Definition of SHM Acceleration (OR force) is proportional to displacement/allow

distance from point (from a fixed point) / a (OR F) = (-)constant x with symbols defined

9

Acceleration (OR force) is in opposite direction to displacement/ Acceleration is towards equilibrium point [Allow towards a fixed point if they have said the displacement is measured from this fixed point] / Signs in equation unambiguously correct, e.g. a (OR F) = - 2 x

9

[Above scheme is the only way to earn 2 marks, but allow 1 mark for motion whose period is independent of amplitude OR motion whose displacement/time graph is sinusoidal]

2(b)(i) Calculation of period Use of T = 2 km / 9 Correct answer [1.1 s] 9 2 e.g. T = 2(0.120 kg / 3.9 N m-1)

= 1.10 s

(ii) Calculation of maximum speed Use of vmax = 2 fxo and f = 1/T 9 Correct answer [0.86 m s-1] 9 2 e.g. f = 1/(1.10 s)

= 0.91 Hz vmax = 2 (0.91 Hz)(0.15 m) = 0.86 m s-1

(iii) Calculation of maximum acceleration Use of amax = (-)(2f)2x0 9 Correct answer [4.9 ms-2] 9 2 e.g. amax = (2 0.91 Hz)2(0.15 m)

= 4.9 m s-2

(iv) Calculation of mass of block Use of T m / Use of T = 2 km / 9 Correct answer [0.19 kg] 9 2 e.g. m = (0.12 kg)(1.4 s /1.1 s)2

= 0.19 kg OR m = (3.9 N m-1)(1.4 s / 2)2 = 0.19 kg

10 [Apply ecf throughout]

38

4.(a)(i) How the bow causes the wave pattern EITHER Bow alternately pulls and releases string (or sticks and slips) 9 Creates travelling wave (OR travelling vibration ) (on string) 9 Wave reflects at the end (OR bounces back) 9 Incident and reflected waves (OR waves travelling in opposite

directions) superpose (OR interfere OR combine) [Dont accept collide]

9

max 3 OR Bow alternately pulls and releases string (or sticks and slips) 9 Produces forced oscillation/acts as a driver/exerts periodic force

[Dont accept makes it vibrate] 9

At a natural frequency of the string 9 Causing resonance (OR large amplitude oscillation) 9 max 3 (ii) Determination of wavelength Use of node to node distance = /2 / recognise diagram shows 2] 9 Correct answer [0.4 m] 9 2 e.g. = 2 0.2 m

= 0.4 m

(iii) Differences between string wave and sound wave Any TWO points from: - String wave is transverse, sound wave is longitudinal / can be

polarised, cant

- String wave is stationary (OR standing), sound wave is travelling (OR progressive) / has nodes and antinodes, doesnt / doesnt transmit energy, does

- The waves have different wavelengths - Sound wave is a vibration of the air, not the string 99 2 [Dont accept travel in different directions / can be seen, cant be seen /

cant be heard, can be heard / travel at different speeds The first two marking points require statements about both waves, e.g.

not just sound waves are longitudinal]

(b) Sketch of the waveform Sinusoidal wave with T = 1 ms

[Zero crossings correct to within half a small square Accept a single cycle]

9

Amplitude 1.6 cm [Correct to within half a small square]

9

2 9

39

5.(a) Conditions for observable interference Any THREE of: Same type of wave / must overlap (OR superpose) / amplitude

large enough to detect / fringes sufficiently far apart to distinguish [Only one of these points should be credited]

(Approximately) same amplitude (OR intensity) Same frequency (OR wavelength) Constant phase difference (OR coherent OR must come from

the same source)

999 3 [Accept two or more points appearing on the same line in the answer

book Dont accept

- must be in phase - must be monochromatic - must have same speed - no other waves present - must have similar frequencies - answers specific to a particular experimental situation, e.g.

comments on slit width or separation]

(b)(i) Experiment description [Marks may be scored on diagram or in text] (Microwave) transmitter, 2 slit barrier and receiver

[Inclusion of a screen loses this mark, but ignore a single slit in front of the transmitter]

9

Barrier, metal sheets [Labels indicating confusion with the light experiment, e.g. slit separations or widths marked as less than 1 mm, lose this mark]

9

Appropriate movement of receiver relevant to diagram [i.e. move in plane perpendicular to slits along a line parallel to the plane of the slits, or round an arc centred between them]

9

3 (ii) Finding the wavelength Locate position P of identified maximum/minimum 1st/2nd/3rd etc.

away from centre 9

Measure distance from each slit to P 9 Difference = OR /2 (consistent with point 1) 9 3 [Accept use of other maxima and corresponding multiple of ] 9

40

6.(a) [Treat parts (i) and (ii) together. Look for any FIVE of the

following points. Each point may appear and be credited in either part (i) or part (ii)]

(i) Light (OR radiation OR photons) releases electrons from cathode

Photon energy is greater than work function / frequency of light > threshold frequency / flight > fo / wavelength of light is shorter than threshold wavelength / < 0

PD slows down the electrons (OR opposes their motion OR creates a potential barrier OR means they need energy to cross the gap)

Electrons have a range of energies / With the PD, fewer (OR not all) have enough (kinetic) energy (OR are fast enough) to cross gap

Fewer electrons reach anode / cross the gap (ii) (At or above Vs) no electrons reach the anode / cross the gap Electrons have a maximum kinetic energy / no electrons have

enough energy (OR are fast enough) to cross

ANY FIVE 99999

[Dont worry about whether the candidate is describing the effect of increasing the reverse p.d. (as the question actually asks), or simply the effect of having a reverse p.d.]

5 (b) Effects on the stopping potential (i) No change 9 (ii) Increases 9 2 [Ignore incorrect reasons accompanying correct statements of the

effect]

7

41

7.(a)(i) Calculation of de Broglie wavelength Use of E = mv2 9 Use of p = mv 9 Use of = h / p 9 = 3.1 10-10 (minimum 2 s.f.) 99 [Use of E = p2/2m earns the first two marks

If the last two marks are not earned, allow 1 mark for v = 2.3 106 or p = 2.1 10-24]

5 e.g. v = (2 x 2.46 10-18 J / 9.11 10-31 kg)

= 2.32 106 m s-1 p = 9.11 10-31 kg 2.32 106 m s-1 = 2.12 10-24 kg m s-1 = 6.63 10-34 J s / 2.12 10-24 kg m s-1 = 3.13 10-10 m

[Reverse argument, calculating the KE from the given wavelength, can earn 4 max: the first three marks, and 1 (only) for the answer 2.7 x 10-18 J, 2 sig fig minimum]

(ii) Region of electromagnetic spectrum X rays 9 1 (iii) Why the electron is suitable Diffraction occurs 9 Wavelength approximately equal to atomic spacing

[Dont accept atomic size] 9

2 (b) Meaning of wave-particle duality QOWC 9 When a particle exhibits wave properties / When a wave exhibits

particle properties [Accept statement about a specific wave/particle, e.g. light or electrons]

9

Electrons behave like particles .. [example] Possible examples: In electric circuits / Can be accelerated in a vacuum tube / Can be deflected by electric (OR magnetic) fields / Can collide / During ionisation / When beta rays are detected in a GM tube

9

Electrons behave as waves .. [example] Possible examples: Can be diffracted / interfere / As stationary waves inside an atom / in tunnelling microscope

9

4 [The examples given should indicate behaviour in which

particle/wave properties are displayed, rather than just mentioning generic properties (e.g. electrons have mass, electrons have charge)]

12

42

43

6735 Unit Test PHY5 1.(i)

Determination of capacitance Correct answer 5 10-4 F

Example of answer

C = VQ =

V 12C 106 3 = 5 10-4 F [= 0.5 mF or 500 F]

9

1

(ii) Feature of graph Area enclosed by the graph and the Q axis/ area under graph [Do not accept just the single word area, nor area of the graph]

9

1

(iii) Calculate energy stored Attempt to determine area / use of e.g. QV or CV2 Correct answer 3.6 10-2 J [= 0.036 J or 36 mJ] [ecf capacitance value from (i) if used in calculation. Also ecf any multiplying factor introduced in error at (i) and applied again here. e.g. 0.5 F at part (i) 36 J here will score both energy calculation marks]

9 9

2

(iv) Add line to graph Steeper straight line. [unless original C value calculated to be greater than 2000 F] Line stops at (6 V, 12 mC) [No ecf] [Mark according to features on graph. Ignore any working shown at (iv)]

9 9

2

6

44

2.(a) Calculation of mans weight

Use of Newtons law equation. Correct answer 684 N [Allow 683 N, or 687 N but only with G = 6.7 10-11] [neither mark for using m 9.81]

Example of answer:

F = 6.67 x 10-11 N m2 kg-2 N 684)m 104.6(

kg 106kg 7026

24

=

9 9

2

(b)(i) Tangential speed at the equator

Use of tr2

Correct answer 465 m s-1

Example of answer:

v = 16

s m 465s 243600

m104.62 =

9 9

2

(ii) Centripetal force Correct answer (2.36 2.45) N [Allow ecf from b(i) for speeds in range 460 ms-1 to 480 ms-1]

Example of answer:

==r

mvF2

m104.6

)s m 470(kg 706

2-1

= 2.42 N

9

1

(iii) Effect on mans weight measurement Measured weight will be less [beware vague it or his weight is less. Not actual weight is less] Difference = resultant force. / Centripetal force = resultant force. / Part of the gravitational force provides the resultant / the centripetal force. [Do not award if increase in measured weight implied] [Correct calculation clearly identifying both points, e.g. Measured weight = 684 2.4 = 681.6N, scores both marks]

Alternative answers: Reaction force = measured weight gets first mark (9) Weight centripetal force = reaction force gets second mark (9) [N.B. No credit for simply comparing values from use of F = mg and F = GMm/r2 , though discrepancy is ~ 2.4 N.]

9 9

2 7

45

3.(a) One difference between E fields of point charge and parallel plates

For parallel plates the field is constant/uniform [not linear]. For point charge the field is radial / reduces [not just varies] with

distance / is 21d . [Both points must be made. Reject simple contrast of uniform with non-uniform, or of parallel field lines with non-parallel ones.]

9

1

(b)(i) Electric field strengths due to each charge

Use of 2rkQ or 24 r

Q

o Correct answer () 3.6 106 N C-1 [or V m-1] Correct answer 2.0 107 N C-1 [Apply unit error once; e.g. 360 N C-1 and 2000 N C-1 scores 2/3. Values of 3.6 10n and 20 10n with no unit score 2/3, for any n. Transposition of r values 1.0 107 N C-1 and 7.2 10-7 N C-1. This pair of answers scores the first mark only]. Example of answer:

][ 2226229

m)10( 3

C )10 ( 2C m N 1099.8

/ same but using 5 cm and 1 C .

9 9 9

3

(ii) Directions of E fields at P due to both charges Two arrows at P, one pointing upwards AND one pointing towards 1 C. [both arrows required for the single mark]

9

1

(c)(i)

(ii)

How to locate the positions of the 2 C charge Single straight line accurately drawn from -1 C through P, intersecting circle at two (diametric) points / Points marked on circle where this line would intersect it. Marking and labelling the points Max (top right) and Min (bottom left) both correctly labelled

9 9

2 4

46

4.(a) Direction of magnetic field

Into end A / through coil left to right [No credit for S or N labelling alone]

9

1

(b)(i) Why a momentary current is formed QoWC Changing current in coil 1 changes field in coil 1 Coil 2 experiences this changing magnetic field (from coil 1) / Flux cutting occurs in coil 2 [Disallow this mark if change in B is attributed to motion] Emf is induced across coil 2 / Emf due to Faradays Law [ or equation form given] / Emf due to flux cutting. Hence (induced) current produced in coil 2 (since in a closed circuit) [dependent mark; award only if previous mark given]. Once current (in coil 1) is steady flux stops changing / no emf is induced / there is no induced current.

9 9 9 9 9 9

Max 5

(ii) Why its direction is Q to P Correct reference to magnetic polarity; i.e. end Q has same polarity as end A, stated or shown anywhere Reference to Lenz's Law./ direction of current ( or emf) in circuit 2 acting to oppose the rise of flux / flux change in circuit 1 [Do not allow acting to oppose motion] [No credit for just direction of current in circuit 2 must be opposite to current direction in circuit 1, nor for idea of emf flowing]

9 9

2

7

47

5.

(a)(i) Directions of B fields B8A is into page and B2A is out of page [not up or down, clockwise and anticlockwise, etc.]

9

1

(ii) Resultant magnetic field, first method

Substitution into aI20 [Allow any I or a from data given]

Both 1.07 10-4 T and 2.67 10-5 T found Their B values subtracted or added correctly, ecf part (a) [Expected difference answer 8.0 10-5 T, sum answer 13.4 10-5 T] [ If a = 0.03 m used in error, 4.0 10-5 T difference value or 6.7 10-5 T sum value, scores 2/3 total. ] Alternative route, first mark: Values of I subtracted or added to give 6A or 10A, ecf part (a), and

substituted into aI20 [Allow any a from data given] (9)

Then either: 8.0 10-5 T found (for 6 A) or 13.4 10-5 T (for 10 A) (scores 99) OR 4.0 10-5 T found (for 6 A) or 6.7 10-5 T (for 10 A) (scores 9)

NB. If difference or sum found clearly contrary to part (a) answers given in terms of into or out of page, max 2 marks. If no clear answer to part (a), or answers there not in terms of into and out of page, fully credit either sum or difference.

9 9 9

3

48

(b)

What happens when the switch is closed Current is down the page/opposite to current direction in wires [Award mark if current direction is correctly shown on diagram] (Resultant) magnetic field is into the page. (ecf their original into or out of page 8A field direction at a(i)) Flemings Left Hand Rule / Magnetic field round foil interacts with / combines with magnetic field(s) round wires. Alternative route for second and third marks only: Foil experiences repulsive force from both wires (9) Force from 8A wire is greater (9) Foil experiences force / moves to the right (Or left, ecf their original into or out of page 8A field direction at a(i)) [Apart from the first mark for current direction in foil, allow no further credit for answers based on electromagnetic induction arguments, nor for those referring to electric fields instead of magnetic fields.]

9 9

9 9

Max 3

7

49

6.(a) Why filament is supplied with thermal energy Work must be done to liberate electrons from the metal structure / metal atoms / metal surface / to overcome electrostatic attraction. / Work function argument. / Electrons must have their kinetic / potential energy increased . 9 [Not simply high temperatures are needed. Reference to thermionic emission gains credit only if explained.]

9

1

(b)(i) Calculation See substitution of correct values of e = 1.6 x 10-19 C and V = 5000V in e V expression for electrical work done See substitution of me = 9.11 x 10-31 kg into mev2 Correct answer 4.2 x 107 m s-1 [Accept 4 107 m s-1] [Reaching v2 = 1.8 x 1015 scores 2/3 even if final answer wrong]

9 9 9

3

(ii) Why electrons do not have identical speeds The electrons are emitted with a range of speeds/range of kinetic energies from the filament. / Electrons from the surface of the filament have more kinetic energy than those emitted from deeper in the metal. / Not all electrons are accelerated from rest. / Electrons are emitted in different directions from the filament. / Electrons within the electron beam interact / repel (to cause acceleration or deceleration).

9

1

5

50

51

6735/2A Practical Test PHY5

Question A

(a)(i) Sensible value of T from 18T 333 15T 33 1 if no unit [2 if whole seconds]

9T 3 2 if not T

3 (ii) D measured correctly to 0.01 cm precision and repeated 3 More than one repeat or zero error checked 3 Correct calculation of T from their data with unit and 2 or 3 s.f. 3 3 (iii) Correct calculation of percentage difference, with mean as

denominator 3

Sensible comment based on difficulties eg damped oscillations 3 Quantitative argument 3 3 (b)(i) d to 0.01 mm or better + unit and 0.30 0.33 mm

[ 0.015 mm if not 30 swg constantan] 3

Repeat and average 3 2 (ii) Sensible values for R1 & R2 to 0.1 precision or better + unit 3 r in range 5.7 6.9 m1 with unit and 2 s.f. [10% if not 30 swg

constantan] 3 2

(iii) Correct identification using d or r 3 Referring to both d and r 3 Idea of eliminating contact resistance or any named systematic

error 3 3

16

52

Sample results

(a)(i) 10 T/s: 6.56, 6.51, 6.56, 6.54 T = 0.654 s (ii) No zero error D/cm: 2.37, 2.37, 2.38 D = 2.37(3) cm m = 48.95 g

T = 81.910)1037.2(

04895.016322

= 0.667 s (iii)

Percentage difference = %1006605.0

654.0667.0

= 2.0% Realistic uncertainty in timing is reaction time of 0.1 s. Percentage uncertainty =

5.61.0 100 1.5%

Uncertainty in diameter is 0.01 cm in 2.37 cm 0.5% So second method is probably more reliable. (b)(i) No zero error. d/mm: 0.31, 0.32, 0.31 d = 0.31 mm R1 = 2.5 R2 = 5.5 r = 2 (5.5 2.5) = 6.0 m-1 (iii) Wire is 30 swg constantan as both d and r are closest to the stated

values for this wire.

Will reduce the error due to contact resistance

53

Question B

(a) Correct units for I and t 3 Q calculated correctly with unit 3 5 sensible values 3 5 good values [Good = 100 C from examiners best line] 3333

33 4 good values 4

3 good values 2

9 (b) Graph Sensible scale [at least grid in each direction] and axes labelled

[ignore units] 3

Plots [ square] 3 Line [thin & straight] 3 3 (c) Large triangle [base 8 cm] 3 Correct calculation + unit 3 1800 2600 F 3 2 s.f. 3 4 16

54

Sample results

(a) V / V I / A t / s Q / C 0.20 15 29 435 0.40 15 59 885 0.60 15 89 1335 0.80 15 119 1785 1.00 15 149 2235

(b) See graph (c)

Gradient = V0.90C2000

C = 2200 F (to 2 s.f.)

55

(b)

56

Question C

(a) Vary L and measure x 3 Vertical rule or set square for x 3 Keep m constant or prisms symmetrical or mass at centre 3 Correct expansion

lnx = rlnL + lnk [allow log]

3

Gradient = r 3 Allow lnx = rlnL + lnk

y = mx + c for last 2 marks

5 (b)(i) Correct units for lnL and lnx 3 lnL and lnx correctly calculated to 2/3 d.p. [Not log]

[Ignore trailing zeros] 3

Graph

Sensible scale [at least grid in each direction] and axis labelled [ignore direction]

3

Plots [1/2 square accuracy] 3 Line [thin and straight] 3 5 (ii) Large triangle [base 8 cm] 3 r in range 2.85 3.05, no unit and 2/3 s.f. 3 2 (c)(i) Measure in several places 3 (ii) Use internal jaws or depth gauge 3 Difference method [can get from part (a)] 3 Sensible precaution for w or x, e.g. zero error, eye level, both sides

of rule 3 4

57

Sample results

(a) Keep mass constant and at centre. Vary L, keeping prisms symmetrical Measure x with a vertical rule [or set square/vernier] If x = kLr

lnx = rlnL + lnk

So a graph of lnx against lnL will be a straight line of gradient r (b)(i) L / mm x / mm ln (L / mm) ln (x / mm)

900 13.7 6.80 2.62 800 9.5 6.68 2.25 700 6.5 6.55 1.87 600 4.2 6.40 1.44 500 2.4 6.21 0.88

See graph (ii)

Gradient = 25.670.600.132.2

r = 2.9

58

(b)

59

6736 Unit Test PHY 6

1 (a)

diamond very good / the best and glass poor thermal conductor numerical support: e.g. kdiamond and kglass quoted (so) diamond conducts heat away from skin 1000 times / much faster than glass

9 9 9

3

(b) Q t AT d Q/t = kAT/d [no mark] units: J s1/W m2, K and m

9 9 9 9

max 3

(c) copper: horizontal line

not below 200 K / linear to origin below ~ 300 K ice: concave curve not above 400 K air: convex curve (as if from origin) not meeting k axis

9

9

9

9

9

9 6

(d) (i) use of pV = nRT substitute 102 103 Pa and 291 K / (273 + 18) K 0.047 m3 /0.0474 m3 [no mark] use of h = 870 (J m3 K1)

0.047 / 0.05 with unit J K 1 [giving 40.9 43.5]

9 9 9 9

4

(ii) 1. air particle of radius r / centre-to-centre distance 2r identified

for ideal gas particles are points 2. affected by intermolecular forces ideal gas molecules only interact in collisions / do not attract each other

9 9 9

9 4

60

(e) (i) sketch: a cylinder labelled r / 2r / d plus length labelled 4r2 / r / r2 more chance of a collision so l

9 9 9 3

(ii) k = hl l = 3k h

substitute: h = 870 J m3 K1 [no mark]

l = 1.6(2) 107 ( m) very much bigger than gas molecule / gas molecule ~ 109 m

9 9

9 3

(f) qowc (do not give if first 3 sentences of paragraph 3 are quoted)

diffusion of particles / gas molecules high k.e. particles swap with low k.e. particles thermal energy spreads / moves from hot to cold heat current same as k.e. / kinetic energy current

9 9 9 9 9

max 4

(g) problems: mention of convection / radiation heat loss at ends of tube / how to measure of hot wire small T of warm water / high s.h.c of water

9 9

9

max 2

61

2 (a) (i)

(ii)

idea that gr2 = constant [it is a number about 4 1014] attempt to read off 3 or more sets of points confirm relationships using 2 or more points [ignore 10n] unit of gr2 (or of Gm): m3 s2 either E, r axes with E positive or E with 1/r2 concave shape with origin marked straight line close to (but not touching) axes at each end from marked origin dissimilar : g attracts only, E attracts and repels / E can be shielded, g cant

9 9 9 9

4 9 9 9 9

4 (b) (i)

(ii)

idea that (g.p.e.) = mass gravitational potential difference 39.2 kJ either: mgh = 39.2 x 103 J or: gh = 24.5 J kg1 h = 2.5 m differ: varying m same e equal levels varying levels / distances same g in car parks, varying E in atoms / car can park

between levels, electron cannot [not ref: to size of forces / to zero of field at infinity / to e zero

up and m zero down ] similar : energy / p.e. quantised

9 9 9 9

4 9 9 9 9

4

Total 16

62

3 (a) (i)

(ii) (iii)

data logger / computer / processor [not light gate] display / monitor /computer / oscilloscope [computer once only] vertical axis is velocity stated use of = s/t 1.2 plus unit m s1 (for both) + and trolley / (air track) glider bouncing between elastic barriers / bands

9 9 9 9 9 9

6 9 9

2

(b) (i) sketch: two coils with a common axis / transformer diagram explanation: current in one coil produces B-field changing current produces changing B-field mention of (magnetic) flux / linking (second coil) so steady e.m.f. in second coil as e.m.f. proportional to Nd/dt / Faradays law

9 9 9 9 9 9

max 5

(ii) use of P = IV = 0.030 W / 30 mW [accept J s1]

9

9 2

Total 15

63

4 (a) (i) A : 224 + 4 seen

Z : 88 + 2 seen [where no numbers on the , give for 224, 88]

9 9

(ii)

either: use of t = ln 2 to give or: use of t/t = n [gives 0.365 y1 or 1.16 108 s1] [gives 2.63] use of N = N0et use of (1/2)n

N/N0 / fraction remaining = 0.16 / 16% / fraction 4/25

9 9 9

(iii)

use anywhere of 1.66 1027 (kg u1) [gives m = 9.8 1030 kg] use of E = c2m [gives E = 8.8 1013 J] use of m2 = 1.6(3) 107 m s1 assume: no recoil of nucleus / all k.e. to alpha / not relativistic

9

9

9

9

9

max 4 (b) (i) 1.6 1019 J used [ gives 217 keV 3.47 1014 J]

E = hc/ = hc/E substitute 6.63 1034 J s and 3.0 108 m s1

= 5.7 / 5.8 1012 (m) (so) photon is / X-ray [no e.c.f.]

9 9 9 9 9

max 4

(ii) 1. diffraction / interference pattern (superposition) / Youngs slits at a crystal / very narrow slits / very fine grating / graphite

2. photo-electric effect / emission of electrons by incident photons / use of GM tube

from metal surface / random clicks

9

9

9

9 4

Total 17

Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG18 4FN Telephone 01623 467467 Fax 01623 450481 Email publications@linneydirect.com January 2007 For more information on Edexcel qualifications, please visit www.edexcel.org.uk/qualifications Alternatively, you can contact Customer Services at www.edexcel.org.uk/ask or on 0870 240 9800 Edexcel Limited. Registered in England and Wales no.4496750 Registered Office: One90 High Holborn, London, WC1V 7BH