1.D

[1]

2.C

[1]

3.D[1]

4.C[1]

5.D[1]

6.B[1]

7.A[1]

8.B[1]

9.B[1]

10.B[1]

11.

density, mass (1)

tough, energy (1)

compressive, tensile (1)

[3]

12.(a)(i)Use of Ek = mv2Correct answer [0.44] (1)

Example of calculation:

1

(ii)Collision energy is more than halved (1), so claim is justified (1)2

(b)Calculation of collision energy [60 kJ] (1)Use of W = Fx (1)Correct answer [500 kN] (1)

Example of calculation:Ek = mv2 = 0.5 1200 (10)2 = 60,000 JW = Fx so 3

(c)Crumple zone increases displacement of car during crash so collision forceis reduced or crumple zone increases collision time and so decreases theacceleration (and force) (1)1

[7]

13.(a)As skydiver speeds up, air resistance will increase (1)Net force on skydiver will decrease, reducing acceleration (1)2

(b)Parachute greatly increases the size of the air resistance (1)When air resistance = weight of skydiver, skydiver is in equilibrium (1)2

(c)Use of as v2 = u2 + 2as or mv2 = mgh (1)Correct answer [7.7 m s1] (1)

Example of calculation:v = = 7.7 m s12

[6]

14.Spelling of technical terms must be correct and the answer must beorganised in a logical sequence (QWC)Mixing of layers leading to eddies/whorls (1)Air circulates around at edge of platform (1)Passenger may be pushed over due to eddies/whorls (1)[3]

15.(a)Proportional / Hookes law limit (1)1(b)Stiffness of sample (1)1(c)Work done / strain energy (1)To stretch (OR strain) wire to fracture (1)2

[4]

16.(a)Use of (1)Correct answer [1.44 106 kg] (1)

Example of calculation:

2

(b)Use of mg = 6rv (1)Correct answer [1.2 m s1] (1)

Example of calculation: (2)2

[4]

17.(a)Reference to free fall whilst bungee is slackIdea of KE increasing as GPE is transformedIdea of work being done against frictional forcesGPE converted into EPE (and KE) once bungee stretchesKE (and GPE) converted into EPE beyond equilibrium pointAt lowest point all of the KE has been converted into EPEMax 4

(b)Use of F = m a (1)Correct answer [6.25 m s2] (1)

Example of calculation:

2

[6]18.(a)Use of s = ut + at2 (1)Correct answer [1.1 s] (1)

Example of calculation:

2

(b)Use of v = u + at (1)Correct answer [1.8 m s1] (1)

Example of calculation:v = u + at = 1.6 1.1 = 1.8 m s12

[4]

19.(a)Length and breadth of the thin cross section to calculate area (1)Digital callipers / micrometer (1)2

(b)Small extensions can be measured accurately (1)Large data set / easy processing of data (1)2

[4]

20.(a)Use of v sin (1)Correct answer [4.2 ms-1] (1)

Example of calculation:v sin = 10 sin 25 = 4.2 m s12

(b)Use of at v = u + at (1)Correct answer [0.43 s] (1)

Example of calculation:v = u + at0 = 4.2 9.8 t = 0.43 s2

(c)Use of or (1)Correct answer [0.90 m] (1)

Example of calculation:

or 2

[6]

21.(a)(i)Assumption: spring obeys Hookes Law (1)Use of F = kx (1)Correct answer [60N] (1)

Example of calculation:

F2 = 6 = 60 N3

(ii)Use of W = Fav.x (1)Correct answer [75J] (1)

Example of calculation:W = Fav.xW = 5 0.5 = 75 J2

(b)(i)Attempt at estimation of area under graph / average force (1)0.5 m extension used (1)Correct answer [53 57J] (1)

Example of calculation:Energy represented by 1 square =10 0.02 = 0.2 J280 squares 0.2 J = 56 J

Treating the area as a large triangleW = 0.5 205 0.5 = 51 J3

(ii)Spelling of technical terms must be correct and the answer must beorganised in a logical sequence (QWC)Energy returned is less than the work done in stretching the cords (1)Energy must be conserved, so internal energy of cords must increase (1)Rubber cords will get warmer (1)3

[11]

22.(a)Idea that no resultant force acts (e.g. forces are balanced / cancel) (1)1(b)Use of w = mg (1)Correct answer [640 N] (1)

Example of calculation:w = mg = 65 9.81 = 638 N2

(c)(i)Tension in rope marked (1)Push from rock face marked (1)Weight marked (1)3

(ii)Use of T = w.sin40 (1)Correct answer [410 N] (1)

Example of calculation:T = w.sin40 = 640 sin40 = 410 N2

(d)Spelling of technical terms must be correct and the answer must beorganised in a logical sequence (QWC)Rigid/stiff exterior to resist deformation under small forces (1)Must undergo plastic deformation under large forces (1)so that collision energy can be absorbed (1)Low density so that helmet is not uncomfortably heavy (1)4

[12]

23.D

[1]

24.C

[1]

25.B[1]

26.A[1]

27.B[1]

28.C[1]

29.B[1]

30.C[1]

31.D[1]

32.(a)D1

(b)Wavelength

Use of v = f (1)Use of f = 1/T (1)Answer T = [0.002 s] (1)[give full credit for candidates who do this in 1 stage T = /v]

Example of answerv = ff = 330 / 0.66T = 1/f = 0.66 / 330T = 0.002 s3

[4]

33.Direction of travel of light is water air (1)Angle of incidence is greater than the critical angle (1)2

[2]

34.(a)Transverse waves oscillate in any direction perpendicular to wave direction (1)Longitudinal waves oscillate in one direction only OR parallel to wavedirection. (1)Polarisation reduces wave intensity by limiting oscillations and wavedirection to only one plane OR limiting oscillations to one direction only. (1)(accept vibrations and answers in terms of an example such as a ropepassing through slits)3

(b)Light source, 2 pieces of polaroid and detector e.g. eye, screen, LED ORlaser, 1 polaroid and detector (1)Rotate one polaroid (1)Intensity of light varies (1)3

[6]

35.Frequency unaltered (1)Wavelength decreases (1)Speed decreases (1)3

[3]

36.(a)Use of sensorEvent happens very quickly OR cannot take readings fast enough (1)Sampling rate: 50+ samples per second (1)2

(b)Initially the temperature is low so current is highResistance of filament increases as temperature increasesCurrent falls to steady value when temperature is constantMaximum heating is when lamp is switched on / when current is highestFilament breaks due to melting caused by temperature riseMax 4

[6]

37.The answer must be clear and the answer must be organised in a logicalsequence (QWC

It was known that X penetrated (1)It was not known that X rays were harmful (1)Doctors died because of too much exposure (1)Lack of shielding (1)New treatments may have unknown side effects (1)Treatments need to be tested / time allowed for side effects to appear (1)Max 4

[4]

38.(a)[1.0 m] (1)1

(b)Ratio of (5 or 6 / 3 ) 60 (1)Answer [f = 100 Hz] (1)2

[3]

39.Use of sin i / sin r = (1)Use of either 80 or 1.33 (1)[r = 48] (1)3

Example of answersin 80 / sin r = 1.33[r = 48]

Both rays refracted towards the normalViolet refracted more than red2

[5]

40.(a)Calculation of energy required by atom (1)Answer [1.8 (eV)] (1)

Example of answer:Energy gained by atom = 13.6 eV 3.4 eV = 10.2 eVKE of electron after collision = 12 eV 10.2 eV = 1.8 eV2

(b)Use of E = hf and c = f (1)Conversion of eV to Joules (1)Answer = [1.22 107 m] (1)

Example of answerE = hf and c = f E = hc/ = (6.63 1034 J s 3 108 m s2) (10.2 eV 1.6 1019 C) = 1.21 107 m3

[5]

41.The answer must be clear, use an appropriate style and be organised in alogical sequence. (QWC)Reference to I = nqvA (1)

For the lampIncreased atomic vibrations reduce the movement of electrons (1)Resistance of lamp increases with temperature (1)

For the thermistorIncreased atomic vibrations again reduce movement of electrons (1)But increase in temperature leads to a large increase in n (1)Overall the resistance of the thermistor decreased with increase intemperature. (1)Max 5

[5]

42.(a)Diffraction is the change in direction of wave or shape orwavefront (1)when the wave passes an obstacle or gap (1)2

(b)The energy of the wave is concentrated into a photon (1)One photon gives all its energy to one electron (1)2

(c)Energy of photon increases as frequency increases OR reference toE = hf (1)Electrons require a certain amount of energy to break free and thiscorresponds to a minimum frequency (1)2

[6]

43.(a)(i)Use of speed = distance over time (1)Distance = 4 cm (1)Answer = [2.7 105 s] (1)

Example of answert = 4 cm 1500 m s1t = 2.7 105 s3

(ii)Use of f = 1/T (1)Answer = [5000 Hz] (1)2

(iii)Time for pulse to return greater than pulse interval (1)All reflections need to reach transducer before next pulse sent. (1)Will result in an inaccurate image. (1) (Max 2)Need to decrease the frequency of the ultrasound. (1) (Max 3)Max 3

(iv)X-rays damage cells/tissue/foetus/baby but ultrasound doesnot (need reference to both X-rays and ultrasound) (1)1

(b)The answer must be clear, use an appropriate style and be organisedin a logical sequence (QWC)

Doppler shift is the change in frequency of a wave when the source or thereceiver is moving (1)Requirement for a continuous set of waves (1)Two transducers required (one to transmit and one to receive) (1)Change in frequency is directly related to the speed of the blood (1)4

[13]

44.(a)Voltmeter is across resistor should be across cell (1)1(b)(i)Plot of graphCheck any three points (award mark if these are correct) (3)Line of best fit3

(ii)e.m.f. = [1.36 1.44 V] (1)1(iii)Attempt to find gradient (1)Answer [0.38 0.42 ] (1)2

(c)Intercept would twice value above (1) (accept numerical value 2 value(b)(ii))Gradient would be twice value above (1) (accept numerical value 2 value(b)(iii))2

[9]

45.C

[1]

46.D[1]

47.C[1]

48.B[1]

49.D[1]

50.A[1]

51.D[1]

52.C[1]

53.A[1]

54.B[1]

55.(a)They act on the same body or do not act on different bodies (1)They are different types of, or they are not the same type of, force(1)2

(b)As the passenger or capsule or wheel has constant speed (1)there is No resultant tangential force (acting on the passenger) (1)2

(c)Friction between seat & person or push of capsule wall on person (1)1

[5]

56.(a)A baryon is a (sub-atomic) particle made up of 3 quarks(1)1(b)n (ddu) (1)p (duu) (1)2

[3]

57.(a)High frequency or high voltage(1)Alternating or square wave voltage(1)2

(b)No electric field inside cylinders (due to shielding) (1)so no force (on electrons) (1)2

(c)As speed increases (along the accelerator), (1)cylinders are made longer so that time in each stays the same(1)2

[6]

58.The answer must be clear, use an appropriate style and be organised in alogical sequence (QWC)-particles fired at (named) metal (film) (1)in a vacuum (1)Most went straight through or suffered small deflections. (1)A few were reflected through large angles or some were reflected alongtheir original path (1)suggesting the mass or charge of the atom was concentrated in a very smallvolume (1)5

[5]

59.(a)(i)Use of E = CV2 (1)Answer [0.158 J] (1)

E = CV2 = 0.5 2200 106 F (12 V)2E = 0.158 J2

(ii)Correct substitution into Ep = mgh (1)Answer 0.75 [75%] (1)

Ep = 0.05 kg 9.8 N kg1 0.24 m [= 0.12 J]Efficiency = 0.12 0.16 J = 0.75 [75%]2

(b)(i)(t = CR) = 2200 106 (F) 16 () = 35.2 (ms) (1)1(ii)Curve starting on I axis but not reaching t axis (1)

I0 = 1.6 V / 16 = 100 mA shown on axis (1)

Curve passing through about 37 mA at t = 35 ms (1)3

(c)(i)The vibrations of the air particles (1)are parallel to the direction of travel of the wave (energy) (1)2

(ii)T = 1/f = 50 ms (1)

Sensible comment related to time constant of 35 ms (1)2

[12]

60.(a)3Li7 + 1p1 = 2He4 + 2He4

completing LHS (1)completing RHS(1)2

(b)(i)Charge (1)(mass/) energy (1)2

(ii)Mass of Li + p = 7.0143 u + 1.0073 u = 8.0216 u (1)Mass of 2 -particles = 2 4.0015 u = 8.0030 u (1)m= 8.0216 u 8.0030 u = 0.0186 u

= 0.0186 1.66 1027 kg = 3.09 1029 kg (1)E= c2m = (3.00 108 m s1)2 3.09 1029 kg = 2.78 1012 J (1)

[Allow ecf from equation]4

(iii) = 1.74 107 eV = 17.4 MeV (1)The incoming proton has an energy of 300 keV = 0.30 MeV (1)So total energy = 17.4 MeV + 0.3 MeV = 17.7 MeV (1)The calculated energy differs by (1)The experiment therefore provides strong evidence for Einsteinsprediction (1)5

[13]

61.(a)Total (linear) momentum of a system is constant, (1)

provided no (resultant) external force acts on the system (1)2

(b)The answer must be clear, use an appropriate style and be organised in alogical sequence

Use of a light gate (1)Use of second light gate (1)

Connected to timer or interface + computer (accept log-it) (1)

Cards on gliders (1)

Measure length of cards (1)Velocity = length time (1)6

(c)Multiplies mass velocity to find at least one momentum (1)

1560 g cm s1 (0.0156 kg m s1) before and after (1)2

[10]

62.(a)(i)Use of A = r2 leading to 0.87 (m) (1)1(ii)Correct use of = 2/t leading to 62.8 (rad s1) (1)1(iii)Correct use of v = r = 55 m s1 [allow use of show that value] (1)1(b)(i)Substitution into p = Av3 (1)

3047 (W) (1)2

(ii)Air is hitting at an angle/all air not stopped by blades (1)Energy changes to heat and sound (1)2

(c)(i)Attempts to find volume per second (A v) (1)

44 kg s1 (1)2

(ii)Use of F = mv/t (1)

F = 610 N (1)2

(d)Recognises that 100 W is produced over 24 hours (1)Estimates if this would fulfil lighting needs for a day(1)Estimates energy used by low energy bulbs in day(1)Conclusion (2)The answer must be clear and be organised in a logical sequence

Example:

The 100 W is an average over the whole day. Most households would uselight bulbs for 6 hours a day in no more than 4 rooms, so this would meanno other energy was needed for lighting.4 low energy bulbs would be 44 W for 6 each hours so would require energyfrom the National grid.

[Accept an argument based on more light bulbs/longer hours that leads tothe opposite conclusion]5

[16]

63.C

[1]

64.B[1]

65.B[1]

66.C[1]

67.C[1]

68.D[1]

69.A[1]

70.B[1]

71.D[1]

72.C[1]

73.(a)as ideal gases do not have forces between molecules so no potential energy (2)2

(b)use of pv = NkTconversion of T to kelvin and answer = 5.8 1022 molecules2

[4]

74.(a)use of counter (+GM tube)determine background count in absence of sourceplace source close to detector and:place sheet of paper between source and counter (or increase distancefrom source 3-7 cm of air) reduces count to background4

(b)alpha radiation only has range of 5 cm in air / wouldnt get through casing (1)1

[5]

75.reason (1)and consequence(1)reason (1)and consequence(1)

2 from:Could be different sizes; So larger one could be further awayCould be different temperatures; So hotter one could be further awayDifferent luminosities; So more luminous one could be further away4

[4]

76.(a)attempt to find gradient eg evidence of triangle on graph or values ofdy/dx (1)value of n (1)2

(b)needs T4 (1)(electrical) power P related to luminosity (1)2

[4]

77.(a)difficult to measure distances to far objects accurately / difficult tomeasure speeds of far objects accurately (1)1

(b)appreciate 1/H is age of universe (1)1(c)fate of universe depends on the density of the universe (1)link between gravity and density (1)Hubble constant is changing due to gravitational forces (1)3

[5]

78.(a)19 protons identified (1)calculation of mass defect (1)Conversion to kg(1)use of E = mc2 (1)divide by 40(1)= 1.37 1012 J(1)

[eg 19 1.007276 = 19.138244 + 21 1.008665 = 40.320209 39.953548 =0.36666

1.66 1027 = 6.087 1028 c2 = 5.5 1011/40 = 1.37 1012 J]6

(b)cannot identify which atom/nucleus will be the next to decaycan estimate the fraction /probability that will decay in a given time /cannot state exactly how many atoms will decay in a set time1

(c)(i)conversion of half life to decay constant[eg = ln 2 / 1.3 109 = 5.3 1010 y1]1

(ii)add both masses to find initial mass (1)use of N = Noet (1)rearrange to make t subject (1)Answer = 4.2 109 years (1)(if 0.84 used instead of 0.94 3 max)[eg total initial mass 0.94t = ln 0.1 / 0.94 / 5.3 1010= 4.2 109]4

[12]

79.(a)(i)Use of F = kx (1)F = ma (1)2

(ii)cf with a = 2x ie 2 = k/m (1)use of T = 2/ to result (1)2

(b)(i)resonance (1)1(ii)natural freq = forcing frequency (1)1(iii)use of c = f (1)answer 9.1 1013 Hz (1)2

(iv)use of T = 1/f eg T = 1.1 1014 (1)rearrange formula (1)to give 550 N/m(1)3

[11]

80.(a)equates F = GMm/ r2 and mv2 /r (1)Use of v = 2r/T (1)Cancel ms To give GMT2 = 42 r3 in any form (1)3

(b)(i)remove constants or cancel G 42 (1)use of idea MT2 / r3 = Constant (1)substitution 27 /36 = Mstar / Msun (1)3

(ii)both will complete orbit in same time period (1)star covers small distance / orbit radius smaller compared to planet (1)2

(c)The answer must be clear, use an appropriate style and be organised in alogical sequence (QWC)Larger planets will move centre of mass towards planet / away from starcentre (1)Star moves faster (1)Doppler shift greater for larger speeds (1)3

[11]

81.(a)top row : 17 1 (14) 4 (1)bottom row: 8 1 7 2 (1)other product helium (1)3

(b)The answer must be clear, use an appropriate style and be organised in alogical sequence (QWC)dead star / no longer any fusion (1)small dense hot / still emitting radiation/light (1)consisting of products of fusion such as carbon / oxygen / nitrogen (1)3

(c)(i)use of 3/2 kT (1)conversion to eV (1)answer [1.3 (keV)] (1)3

(ii)gravitational force does work on hydrogen (1)increases internal energy of gas (1)2

(d)The answer must be clear, use an appropriate style and be organised in alogical sequence (QWC)A standard candle (in astronomical terms) produces a fixed amount of light/luminosity (1)Quantity of hydrogen (1)and fusion temperature (1) must be similar for various novae.3

[14]

82.Add missing information

For three correct responses in the vector or scalar column (1)For the base unit column:3 correct responses (2)2 correct responses (1)3

QuantityBase unitVector or scalar

mvector

kg m2 s2scalar

kg m2 s3scalar

kg m s1vector

[3]

83.(a)(i)Describe motionConstant / uniform acceleration or (acceleration of) 15 m s2 (1)

(Followed by) constant / uniform speed / velocity (of90 m s1) (1)2

(ii)Show that distance is approximately 800 mAny attempt to measure area under graph or select appropriateequations of motion required to determine total distance (1)

Correct expression or value for the area under the graph betweeneither 0-4 s [240 m] or 4-10 s [540 m] (1)

Answer : 780 (m) (1)

Eg distance= 60 m s1 4 s + 90 m s1 6 s

= 240 m + 540 m

= 780 (m)

Eg distance in first 4 ss = 4 s = 240 mDistance in final 6 ss = ut = 90 m s1 6 s = 540 mTotal distance = 240 m + 540 m = 780 (m)3

(b)Sketch graphGraph starts at 760 m 800 m/their value and initially showsdistance from finishing line decreasing with time (1)The next two marks are consequent on this first mark being awarded

Curve with increasing negative gradient followed by straight line (1)

Graph shows a straight line beginning at coordinate (4 s, 540 m)and finishes at coordinate (10 s, 0 m) (1)3

[8]

84.(a)(i)Give expressionW = R + F (1)1

(ii)Complete statements. surface / ground (1). Earth (s mass) [Only accept this answer] (1). gardener(s hands) / hand(s) (1)3

(b)(i)Add to diagramLine inclined to the vertical pointing to the left and upwards (1)1

(ii)Explain change in direction and magnitudeThe force (at X) will have a magnitude greater than For the force (at X) must increase. (1)

This is because the wheelbarrow / it has to be lifted /tilted/ supported/ held up (by the vertical component) (1)

And also because the wheelbarrow / it has to be moved(forward by the horizontal component) (1)3

[8]

85.(a)Show that rate of decay of radium is about 7 1013 BqPower divided by alpha particle energy (1)

Answer [(7.1 7.2) 1013 (Bq)] (1)

[Give 2 marks for reverse argument ie7 1013 Bq 7.65 1013 J (1)(53.5 53.6) (W) (1)]Eg Rate of decay =

= 7.19 1013 (Bq)2

(b)Show that decay constant is about 1.4 1011 s1Use of = (1)Answer [(1.35 1.36) 1011 (s1)] (1)

Eg = = 1.35 1011 (s1)2

(c)The number of radium 226 nucleiUse of A = N (1)

Answer [(5.0 5.4) 1024] (1)Eg 7.19 1013 Bq = 1.35 1011 s1 NN = 5.33 10242

(d)The mass of radiumDivides number of radium 226 nuclei by 6.02 1023 andmultiplies by 226 (1)

Answer [1870 2040 g]

Eg Mass of radium = 226 g = 2008 g2

(e)Why mass would produce more than 50 WThe (daughter) nuclei (radon) formed as a result of the decay ofradium are themselves a source of (alpha)radiation / energy (1)Also accept(having emitted alpha) the nucleus[allowsample/radium/atom] (maybe left excited andtherefore also) emits gammaAlso accept(daughter) nucle(us)(i) recoil releasing (thermal) energyDo not acceptNucleus may emit more than one alpha particleNucleus may also emit beta particle1

[9]

86.(a)Paths of alpha particlesPath A drawn less deflected than B (1)

Path A drawn as a straight line (1)2

(b)(i)Why alpha source inside containerAlpha would be absorbed by [accept would not get through]container (material) (1)1

(ii)Why the same kinetic energy?EitherTo restrict observation to two variables / closeness of approachand deflectionor so that speed / velocity / (kinetic) energy does not havean effect (on the observation / deflection /results /contact time)1

(iii)Why an evacuated container?Eitherso that alphas do not get absorbed by / collide with / getdeflected by / stopped by / scattered by / get in theway of / ionise / lose energy to atoms / molecules (ofair) [Do not accept particles of the air]

or so that all alphas reach the foil with the same (kinetic) energy1

[5]

87.(a)Diode or LED (1)1

(b)(i)Use of R = V / I current between 75 and 90 ignoring powers of 10 (1)answer 6.7 8.0 (1)

Example of answerR = 0.60 V (85 103) AR = 7.06 2

(ii)Infinite OR very high OR 1

(c)ANY ONERectification / AC to DC / DC supply [not DC appliances]Preventing earth leakageStabilising power outputTo protect componentsA named use of LED if linked to LED as component in (a)(egcalculator display / torch)A voltage controlled switch(Allow current in only one direction)1

[5]

88.(a)Resistivity definition

Resistivity = resistance (1)

cross sectional area / length (1)

= RA/l with symbols defined scores 2/2equation as above without symbols defined scores equation given as R = l/A with symbols defined scores 1/2

(1st mark is for linking resistivity to resistance with some other terms)2

(b)(i)Resistance calculationConverts kW to W (1)Use of P = V2/R OR P = VI and V = IR (1)Resistance = 53 (1)

Example of answerR = (230 V)2 1000 WR = 53 3

(ii)Length calculationRecall R = l/A (1)Correct substitution of values (1)Length = 6.3 m (accept 6.2 m) (1)ecf value of R

Example of answerl = (52.9 1.3 107 m2) (1.1 106 m)l = 6.3 m3

(iii)Proportion methodIdentifies a smaller diameter is needed (1)Diameter = 0.29 mm (1)ORCalculation methodUse of formula with l = half their value in (b)(ii) (1)Diameter = 0.29 mm (1)(Ecf a wrong formula from part ii for full credit)

Example of answerdnew = 0.41 mm 2dnew = 0.29 mm2

[10]

89.(a)Definition of E.M.F.Energy (conversion) or work done (1)Per unit charge (1)[work done/coulomb 1/2, energy given to a charge 1/2, energygiven to a charge of a coulomb 2/2]ORORE = W/Q (1)E = P/ISymbols defined (1)Symbolsdefined(E = 1 J/C scores 1)(E = 1 W/A scores 1)

((Terminal) potential difference when no current is drawn 1/2)2

(b)(i)Internal resistance calculationAttempt to find current (1)Pd across r = 0.2 V (1)r = 0.36 () (1)[You must follow through the working, I have seen incorrectmethods getting 0.36 ]

Example of answerI = 2.8 V 5.0 r = (3.0 2.8) V 0.56 A = 0.36 3

(ii)Combined resistanceUse of parallel resistor formula (1)Resistance = 3.3 [accept 3 1/3 but not 10/3] (1)2

(iii)Voltmeter reading(ecf bii)Current calculation using 3 V with either 3.3 or 3.7 (1)Total resistance = 3.7 [accept 3.66 to 3.73 ]OR use of V = E Ir (1)Voltmeter reading = 2.7 V (1)

ORPotential divider method, ratio of resistors with 3.7 on bottom (1)Multiplied by 3.0 V (1)2.7 V (1)

Example of answerRtotal = 3.7 I = 3 V 3.7 = 0.81 AVvoltmeter = 3.3 0.81 A = 2.7 V3

(c)Ideal voltmeterIdeal voltmeter has infinite resistance OR extremely high resistanceOR highest possible R OR much larger resistance than that ofcomponent it is connected across OR quotes value > 1 M (1)

Current through voltmeter is zero (negligible) OR doesnt reduce theresistance of the circuit OR doesnt reduce the p.d. it is meant tobe measuring. (1)2

[12]

90.(a)Circuit diagramPotentiometer correctly connected i.e potential divider circuit (1)Ammeter in series and voltmeter in parallel with bulb (1)

(light bulb in series with resistance can score second mark only)2

(b)(i)Graph+I, +V quadrant; curve through origin with decreasing gradient (1)

[do not give this mark if curve becomes flat and then starts goingdown i.e. it has a hook]

I, V quadrant reasonably accurate rotation of +I,+V quadrant (1)2

(ii)Shape of graphAs current/voltage increases, temperature of the lamp increases /lamp heats up (1)Leading to increase in resistance of lamp (1)Rate of increase in current decreases OR equal increases in Vlead to smaller increases in I (1)Qowc (1)

Ecf if a straight line graph is drawn max 3R constant (1)V I (1)Qowc (1)4

[8]

91.(a)Absolute zero of temperature(Temperature at which) pressure / volume (of a gas) is zero. (1)OR(Temperature at which) the kinetic energy of the molecules is zero)1

(b)(i)Number of moles show that calculation

Recall pV = nRT (1)Addition of air pressure (1)Conversion to kelvin (1)Number of moles = 0.52 (mol) (1)

Reverse calculations using n = 0.5 to arrive at one of the othervalues can score maximum 3

Example of answer

n = 0.52 mol4

(ii)Mass of airMass = 1.5 102 kg (1)

Example of answermass = 0.52 mol 0.029 kg mol1 = 0.015 kg1

(iii)Temperature calculation

Use of P1/T1 = P2/T2 (1)Correct P2 1.6 105 Pa (1)Lowest temperature = 216 K (57 C) (1)

ORUse of pV = nRT (must see correct value of R) (1)Correct P2 1.6 105 Pa (1)Lowest temp 215K 223K (58 to 50 C) (1)

Example of answer

T2 = 216 K3

[9]

92.Core remnant stars

All core remnants ticked AND no main sequence (1)

< 1.4 M column: White dwarf only (1)

> 2.5 M column: Black hole only (1)3

[3]

93.(i)Hydrogen burning

Quality of written communication (1)

Nuclear fusion reaction [accept nuclei, nucleus, fusing] (1)

Hydrogen / deuterium /protons turn into He [penalise contradictions,e.g. molecules atoms; accept symbols ] (1)

Release of energy (1)4(ii)Sun as red giant calculation

Attempted use of L = T4A (accept r substituted as A) (1)

A = 4 r2 [or A r2 if ratios calculated directly] (1)

3.85 1026 (W) or 1.13 1030 (W) [or substitution as ratio] (1)

2930 [accept 2900 2940] (1)

L = T4A = 4 T4r2Lbefore= 4 5.67 108 W m2 K4 (5780 K)4 (6.96 108 m)2

= 3.85 1026 WLafter= 4 5.67 108 W m2 K4 (3160 K)4 (1.26 1011 m)2

= 1.13 1030 WHence ratio = 1.13 1030 W 3.85 1026 W = 29304

(iii)H-R diagram plots

X at 100 on main sequence [ 1 mm by eye] AND between 5000 K andcentre of 5000 10 000 K box (1)

Y above and to right of actual X (1)

Attempt to plot Y at 3160 K [between 5000 K and 2500 K] (1)

Attempt to plot Y between 103 L and 104 L [ecf] (1)4

[12]

94.(i)Sun as white dwarf

Any 2 [comparative statements] of

Higher temperature / hotterLower luminosity [accept Power, not E or I]No fusion in core [or equivalent; not just not on main sequence]More dense (1) + (1)2

(ii)Future of white dwarf

Cools / T decreases (1)

Dims / fades / correct colour change [not brown dwarf] / Luminositydecreases [accept intensity here] (1)2

[4]

95.(i)Distance to Sirius

Substitution in v t /s [ignore 8.6, accept 365 or 365 days] (1)

8.1 1016 (m) [8.13, 8.14] (1)

d= v t= 8.6 3.00 108 m s1 (60 60 24 365) s= 8.1 1016 m2

(ii)Sirius A intensity calculation

Use of I = L / 4 D2 (1)

Correct substitution (1)

1.2 107 W m2 [1.20 1.24] (1)

I= L / 4 D2= 1.0 1028 W / 4 (8.1 1016 m)2= 1.2 107 W m23

(iii)Mass rate conversion

E = m c2 seen [or implied] (1)

Correct substitution (1)

1.1 1011 kg (s1) (1)

1.0 1028 W = 1.0 1028 J s1m= E / c2= 1.0 1028 J / (3.00 108 m s1)2= 1.1 1011 kg3

(iv)Peak wavelength calculation

Use of Wiens law (1)

2.93 107 m (1)

max= 2.90 103 m K / 9900 K= 2.93 107 m2

[10]

96.(i)Fence wire cross-section

Use of r2 and 103 m (1)

4.9 106 (m2) [do not accept m] (1)

A= r2= (0.5 2.50 103)22

(ii)Stress calculation

Substitution: 1500 N / 4.9 [or 5] 106 m2 (1)

310 MPa [accept 300, ecf] (1)

= F / A= 1500 N / 4.9 106 m2= 3.1 108 Pa2

(iii)Extension calculation

E = / and = l / l (or E = F l / A l) (1)

Substitution in E = / and = l / l [or in E = F l / A l, ecf,ignore 10n] (1)

0.048 (m) [ecf] (1)

48 mm [accept 47 49 mm, bald answer scores 4/4] (1)

E= F l / A ll= (1500 N 33 m) / (210 109 Pa 4.9 106 m2)= 0.048 m = 48 mm4

[8]

97.(i)Young modulus experiment

(G) clamp [vice], wire, pulley, mass / weight / load

three correct (1)

all four correct (1)2(ii)Labelling of l

Accurate indication of l [to 1 mm] (1)

Length 2 m to 6 m (1)2

(iii)Additional apparatus

Micrometer (screw gauge) / (digital) callipers (1)

Ruler or similar [e.g. tape measure, metre stick] (1)2

[6]

98.Particle classification

Neutron: baryon and hadron (1)

Neutrino: lepton (1)

Muon: lepton (1)3

[3]

99.(i)Binding energy

Energy required to separate a nucleus into nucleons (1)1(ii)8n + 6p (1)

Substitution / m = 0.1098 u (1)

Multiply by 930 [only, or E = m c2 route] (1)

102 MeV [or 103 MeV] (1)

m = (6 1.007 28 u) + (8 1.008 67 u) 14.003 24 u = 0.1098 uE = 0.1098 u 930 MeV/u = 102 MeV4

(iii)More stable isotope

Binding energy per nucleon attempted (1)

7.4 (MeV) and 7.3 (MeV) [accept 7.1, ecf] (1)

Hence carbon12 [based on two values, ecf] (1)

BE / A (14C) = 102 MeV / 14 = 7.3 MeVBE / A (12C) = 89 MeV / 12 = 7.4 MeV3

[8]

100.(i)Conservation laws

First reaction, Q: 0 + 0 1 + 1 (1)

Second reaction B: 1 = 1 + 0 AND Q: 1 = 1 + 0 (1)

Hence only decay possible [based on B and Q conservation for thisdecay, accept simple ticks and crosses] (1)3

(ii)Quark charges

Use of sss = 1 to show s = (1)

Hence correct working (from baryons) to show u = and d = (1)2

[5]

101.(a)(i)Why speed is unchanged

Force/Weight [not acceleration] is perpendicular tovelocity/motion/direction of travel/instantaneous displacement[not speed]OR no component of force/weight in direction of velocity etc (1)

No work is doneOR No acceleration in the direction of motion (1)2

(ii)Why it accelerates

Direction (of motion) is changing (1)

Acceleration linked to a change in velocity (1)2

(b)Speed of satellite

Use of a = v2/r (1)

Correct answer [3.8 to 4.0 103 m s1] (1)

Example calculation:v = (2.7 107 m 0.56 m s2)

[Allow 1 mark for = 1.4 104 rad s1]2

[6]

102.(a)(i)Demonstrating the stationary wave

Move microphone between speaker and wall OR perpendicular to wallOR left to right OR towards the wall [could be shown by labelledarrow added to diagram] (1)

Oscilloscope/trace shows sequence of maxima and minima (1)2

(ii)How nodes and antinodes are produced

Superposition/combination/interference/overlapping/crossingof emitted/incident/initial and reflected waves (1)

Antinodes: waves (always) in phase OR reference to coincidenceof two compressions/rarefactions/peaks/troughs /maxima/minima,hence constructive interference/reinforcement (1)

Nodes: waves (always) in antiphase/exactly out of phase ORcompressions coincide with rarefactions etc, hence destructiveinterference / cancellation (1)3

(iii)Measuring the speed of sound

Measure separation between (adjacent) nodes / antinodes anddouble to get /this is [not between peaks and troughs] (1)

Frequency known from/produced by signal generator ORmeasured on CRO / by digital frequency meter (1)

Detail on measurement of wavelength OR frequencye.g. measure several [if a number is specified then 3] nodespacings and divide by the number [not one several times]OR measure several (3) periods on CRO and divide by the numberOR adjust cro so only one full wave on screen (1)

Use v (allow c) = f4

(b)(i)Application to concert hall

Little or no sound /amplitudeOR you may be sat at a node (1)(ii)Sensible reason

Examples:Reflected wave not as strong as incident waveOR walls are covered to reduce reflectionsOR waves arrive from elsewhere [reflections/different speakers]OR such positions depend on wavelength / frequency (1)2

[11]

103.(a)(i)Amplitude and frequency

0.17 m (1)

0.8(3) Hz or s1 (1)2

(ii)Maximum velocity

Use of vmax = 2fx0 (1)

Correct answer (1)

Example calculation:vmax = 2 0.83 Hz 0.17 m

ORUse of maximum gradient of h versus t graph

Answer to 2 sig fig minimum2

(iii)Velocity-time graph

Wave from origin, period 1.2 s (1)

Inverted sine wave with scale on velocity axis &initial peak value0.9 m s1 (1)2

(b)(i)Definition of SHM

Acceleration / resultant force proportional to displacementOR Acceleration / resultant force proportional to distance from afixed point [not just distance from equilibrium but distance fromequilibrium position is acceptable]OR a = () constant x [with a and x defined]OR F = () constant x [with F and x defined] (1)

Acceleration /resultant force directed towards the fixed point / inopposite direction (to displacement)OR negative sign in equation explained [e.g a and x in oppositedirections] (1)2

(ii)Verifying SHM

Read off h value and use it to get displacement (1)[only penalise the first mark if h used for displacement throughout]Plot acceleration-displacement graphOR calculate ratios eg ax (1)Straight line through the originOR check ratios to see if

constant (1)Negative gradient / observe accelerationOR constant is negative (1)and displacement have opposite signs

OR

Use x = xocos(2ft) for a range of tOR Read off h and get x (1)Use values of xo and f from part (a)OR Use a = (2f)2x for range of x (1)Add equilibrium value to x to get hOR Use value of f from part (a) (1)If results agree with values of h (or a) from graph it is SHM (1)4

[12]

104.(a)(i)Line B

Knot T at 2.4 m [ small square, no label needed] (1)(ii)Knots Q, R, S at 0.6, 1.2, 1.8 m [ small square, no labels needed][ecf from wrong position of knot T i.e. Q at T, R at T & R at T](1)2

(b)How model represents the Universe and its behaviour

Knots/letters/points represent galaxies (1)

Reference to expansion of Universe / galaxies moving apart[NOT galaxies move away and stay same distance apart] (1)2

(c)How model illustrates Hubbles law

Stating or showing velocities are different for 2 of the knots (1)[Shown by either calculating speeds or comparing distances movedbetween diagrams A and B]

Calculation of velocity for at least 2 of the knots [other than T] (1)

Use of their data to show speed (of knot) distance (from P) (1)Examples:determine values of vd [allow vd]sketch graph of v against d [allow v against d]3

(d)Defects of the model

Any 2 sensible points (1)(1)

Examples:Galaxies are not evenly spacedInitial spacing of knots is not zeroNo force pulling galaxies/Universe apartRate of expansion of Universe OR speed of galaxies increasing/not constant [not speed decreasing]Relative sizes of knot and spacing are unrealisticUniverse is 3 dimensional/galaxies are not in a straight line2

[9]

105.(a)Meaning of statement

(5.89 1019 J / work function) is the energy needed to remove anelectron [allow electrons] from the (magnesium) surface/plate

Consequent markMinimum energy stated or indicated in some way [e.g. at least /or more] (1)2

(b)(i)Calculation of time

Use of P = IA (1)

Use of E = Pt (1)

[use of E = IAt scores both marks]

Correct answer [210 (s), 2 sig fig minimum, no u.e.] (1)[Reverse argument for calculation leading to either intensity,energy or area gets maximum 2 marks]

Example calculation:t = (5.89 1019 J)/(0.035 W m2 8 1020 m2)3

(ii)How wave-particle duality explains immediate photoemission

QOWC (1)

Photon energy is hf / depends on frequency / depends on wavelength (1)

One/Each photon ejects one/an electron (1)

The (photo)electron is ejected at once/immediately (1)[not just photoemission is immediate]4[9]

106.(a)(i)GMS/R2 (*)

(ii)GME/r2 (*) (1)1

(*) (symbols must be as given in the Q, though allow lower case m)(b)(i)Evidence of equating of GMS/R2 and GME/r2 (ecf from part a) (1)

Correct answer 570 580 (1)

Example of answer:

2

(ii)1.5 108 km 1/601 [ignore powers of 10 in distance value] (1)

Correct answer 2.5 2.6 105 km (or 2.5 2.6 108 m) (1)2(c)Letter L on or against line to left of point P(coming within one Earth radius of dotted line) (1)

Reason*:[*Consequent marks; allow only if L position correct or not shown]

Reference to centripetal force/centripetal acceleration/(net) force towards Sun (1)

Force due to Sun must be > force due to Earth (1)3

[8]

107.(a)(i)W = QV (1)

Correct answer 3.2 nJ [3.2 109 J, etc.] (1)

Example of answer:

W = QV = 0.8 109 C 4.0 V = 3.2 109 J2

(ii)+0.8 (nC) on top plate and 0.8 (nC) on bottom plate (1)(both needed)1

(b)Statement (E =) Area or (E =) QV (1)

See calculation 4.0 0.8 or base height (1)

ORC found from graph (1)Use of W = CV2 (1)

Example of answer:

W = CV2 = = 1.6 109 J2

(c)(i)Correct answer 0.2 nC (1)1

(ii)Graph is straight and through origin (1)

ends at 3.0V and their Q (1)2

(iii)Attempt to use C = Q/V or C = Q/V (1)

Correct answer 0.067 nF / 67 pF (1)

Example of answer:

= 6.7 1011 F2

[10]

108.(a)(i)1.2 keV = 1.2 103 1.6 1019 JORUse of eV with e as 1.6 1019 C and V as 1200 V (1)

Use of (mev2) with me as 9.1(1) 1031 kg. (1)

Correct answer 2.0 2.1 107 m s1 (1)3

(ii)1200 8/100 = 96 (eV delivered per electron) (1)96/2.4 = 40 (1)

Or2.4 100/8 = 30 (incident eV needed per photon) (1)1200/30 = 40 (1)

Or1200 / 2.4 = 500 (photons per electron, ideally) (1)500 (8/100) = 40 (1)2

(b)Electrons on screen repel electrons in beam / force opposeselectron motion/decelerating force (1)

Electrons (in beam) decelerated /slowed /velocity reduced/ work done by electrons (against force) (1)

Electron (kinetic) energy reduced (not shared) (1)

Fewer photons (per electron, stated or implied) (1)

Trace less bright (1)

QoWC (1)Max 4

[9]

109.(a)Scale interval is 0.1 (V) (1)1

(b)(i)Use of = ()N/t (1)

Correct answer 9.6 107 (Wb) / 0.96 (Wb) [ignore +/] (1)

Example of answer:

= = 0.12 V = 9.6 107 Wb2

(ii)Use of or or flux = BA, or B = t/NA (1)

Correct answer 0.012 T / 0.013 T (1)

Example of answer:

= BA

B = = 0.012 T[N.B. = 0.96 Wb 0.012T, = 1Wb 0.013T]2

[5]

e t110.(a)pair of values of k.e. and 2 read from graph / gradient (1)

2 > 5 1016 m s2 (1)

mp = 1.62 1.69 1027 (kg) to 3 s.f. (1)3

(b)(i)(values 1.3 1.7, 3.1 3.5, 6.0 6.5) any two correct (1)(1)(ii)E = c2m / E = mc2 (1)

one value for m ( 1028 kg) (1)

use of mp from (i) [no mark]

one value of m/mp : about 10%, 20%, 40% (1)5

[8]

111.(i)appreciation that area of (first) rectangle / at gives speed (1)

accel = (3 m s2)(8 s) / 30 small squares each worth 0.8 m s1 (1)

24 m s1 (1)

(ii)appreciation that area of second is of same area as first /decel = (4 m s2)(6 s) [negative idea not needed] (1)4

(iii)use of P = IV / E = IVt (1)

use of P = F / E = Ft (1)

(3000 N) = (96 A)(750 V) / equating the Ps or Es (1)

= 24 m s13

[7]

112.(a)Comment on use of weighing

Clear statement correctly identifying weight or mass (or theirunits)e.g. kg a unit of mass, not weight (1)1

(b)Calculation to check statement

Use of equation of motion to show time or distance (1)Answer to 2 sig figs [120 m or 4.5 s] [no ue] (1)

Example of calculation:

s = ut + at2s = 0 + 9.81 m s2 (5s)2OR100 = 0 + 9.81 m s2 t2s = 123 mORt = 4.5 s2

(c)Calculation of kinetic energyEitherUse of equation(s) of motion which allow(s) v2 or v to be found (1)Recall of ke = mv2 (1)Answer [69 000 J] (1)

ORRecall of Ep = mgh (1)Substitution (1)Answer [69 000 J] (1)

Example of calculation:2 = u2 + 2as2 = 0 + 2 9.81 m s2 100 m2 = 1962 m2 s2ke = 1/2 mv2= 69 000 J (68 670 J)

ORgpe = mghgpe lost = 70 kg 9.81 N kg1 100 mgpe lost = 69 000 J (68 670 J)[so ke = 69 000 J because ke gained = gpe lost]3

[6]

113.(a)(i)Condition for reflection

Angle of incidence greater than critical angle [accept i > c] (1)1

(ii)Description of path of ray

Any two from:Ray refracted at A and CDescription of direction changes at A and CTotal internal reflection at B (1)(1)2

(b)(i)Things wrong with the diagram

Angle of refraction cant be 0 / refracted too much (1)

No refraction on emergence from prism (1)[Allow 1 mark for correct reference to partial reflection]2

(ii)Corrected diagramemergent ray roughly parallel to the rest of the emergent rays (1)direction of refraction first surface correct (1)direction of refraction second surface correct (1)3

[8]

114.(a)Formula for C6

v = u + at OR v = 10.7 (9.81 0.2) [units need not be given]OR C6 = C5 9.81*A6 (1)1

(b)Explain B5 to B16 constantg affects vertical motion only / no horizontal force (1)1

(c)Significance of negative valuesThe ball moving downwards (1)1

(d)(i)Completion of diagramVertical arrow has 6.8 added, horizontal arrow has 10.7 added (1)1

(ii)Calculation of velocity at time t = 0.4 s

Use of Pythagoras

Answer for magnitude of v [12.7 m s1] [ecf from diagram]

Use of trigonometrical function [ecf from magnitude]

Answer for direction [32.4] [ecf from diagram]

Example of answer:v2 = (6.8 m s1)2 + (10.7 m s1)2v = 12.7 m s1tan = 6.8 m s1 10.7 m s1 = 32.4

[For scale drawing components drawn correctly to scale(1),resultant shown correctly (1), answer for v 0.5 m s1 (1), angle to 2(1)]4

(e)(i)Calculation of components for new angle

Answer for vertical component [8.7 m s1] (1)Answer for horizontal component [12.5 m s1] (1)

[1 mark only if answers reversed]

Example of answer:vertical component = v sin = 15.2 m s1 sin 35 = 8.7 m s1horizontal component = v cos = 15.2 m s1 cos35 = 12.5 m s12

(ii)Suggest reason for greater distance

Examples greater horizontal component of velocity; easier tothrow at higher speed closer to the horizontal; launching fromabove ground level affects the range; force applied for longer;more force can be applied (1)1

[11]

115.(a)Show that heat energy supplied at about 500 W

Recall of power = energy/time (1)

Answer to 2 sig figs [470 [W]] [no ue] (1)

Example of calculation:

power = energy/time= 1.63 105 J/ 347 s= 470 W2

(b)(i)Show that heat energy gained is about 1 105 J

Use of Q = mc (1)

Correct answer [1.4 105 [J]] [no ue] (1)

Example of calculation:

Q = mc= 0.44 kg 3800 J kg1 C1 (96 C 12 C)= 1.4 105 J2

(ii)Calculate the time taken to reach 96 C

Use of time = energy/power (1)

Correct answer [300 s] (1)

Example of calculation:

time = energy/power= 1.4 105 J/ 470 W= 299 s2

(c)(i)Explain why it might take longer

Heat supplied to milk at a lower rate / expansion onmechanism of heat loss /destination of heat lost (1)1

(ii)Suggest why time the same

Power calculation includes a heat loss factor / rate of heat gain thesame as for water / appropriate mechanism to reduce heat loss(Allow 1 for heat losses already taken into account whenwarming the water) (1)1

[8]

116.(a)Calculation of adaptors input

Recall of: power = IV (1)

Correct answer [0.01 A] (1)

Example of calculation:

power = IVI = P/V = 25 W / 230 V= 0.01 A2

(b)(i)Explain why VA is a unit of power

Power = voltage current so unit = volt amp (1)1

(ii)Calculation of efficiency of adaptor

Use of efficiency equation (1)

Correct answer [24%] (1)

Example of calculation:

efficiency = (0.6 VA / 2.5 W) 100%= 24 % [0.24]2

(iii)Reason for efficiency less than 100%

Resistance (accept explanations beyond spec, e.g. eddy currents) (1)

hence heat loss to surroundings (1)2(c)(i)Calculation of charge

Recall of: Q = It (1)

Correct answer [4000 C] (1)

Example of calculation:

Q = It= 0.2 A 6 h= 0.2 A (6 60 60) s= 4000 C (4320 C)2

(ii)Calculation of work done

Recall of: W = QV OR Recall of W = Pt (1)

Correct substitution (1)

Correct answer [13 000 J] (1)

Example of calculation:

W = QVW = 4320 C 3 V [ecf]= 13 000 J (12 960 J)ORW = PtW = 0.6 W 6 hW = 0.6 W (6 60 60) s= 13 000 J3[12]

117.(a)(i)Add standing waves to diagrams

Mark for each correct diagram (1)(1)2

(ii)Mark place with largest amplitude of oscillation

antinode marked [allow clear indication near centre of wave otherthan an X, allow correct antinode shown on diagrams B or C] (1)1

(iii)Name of place marked

(Displacement) Antinode [allow ecf from (a) (ii)] (1)1

(b)(i)Calculation of wavelength

Correct answer [5.6 m]

Example of calculation:= 2 2.8 m= 5.6 m (1)1

(ii)Calculation of frequency

Recall of v = f (1)

Correct answer [59 Hz] [ecf] (1)

Example of calculation:v = f f = 330 m s1 / 5.6 m= 58.9 Hz2

(c)(i)Explanation of difference in sound

as the room has a standing wave for this frequency / wavelength /it is the fundamental frequency(allow relevant references to resonance) (1)1

(ii)Suggest another frequency with explanation

Appropriate frequency [a multiple of 59 Hz] [ecf] (1)

Wavelength 1/2, 1/3 etc (stated or used) (1)2

(d)Explain change in frequencies

wavelengths (of standing waves) bigger / f = v/2l (1)

hence frequencies smaller/lower (1)2[12]

118.(a)Blue light:Wavelength / frequency / (photon) energy1

(b)(i)Frequency:Conversion of either value of eV to JoulesUse of f = E / hCorrect frequency range [4.8 1014 8.2 1014 Hz or range =3.4 1014 Hz][no penalty for rounding errors]

eg.2 eV = 2 1.6x 1019 = 3.2 1019 J= 6.63 1034 ff = 4.8 1014 Hz3.4 eV = 3.4 1.6 1019 = 5.4 1019 Jf = 8.2 1014 Hz3

(ii)Diagrams:Downward arrow from top to bottom levelOn larger energy gap diagram2

(c)(i)Resistivity drop:Less heating / less energy lost / greater efficiency / lowervoltage needed / less power lost1

(ii)Resistance:Recall of R = L/AUse of R = L/ACorrect answer [80()] [allow 8084 () for rounding errors]

Eg.R = (2 102 5.0 103) / (3.0 103 4.0 104)= 83 3

[10]

119.(a)(i)Type of behaviour:PlasticCorrect definition of circled word:Ductile: can be pulled into a long thin shapeElastic: returns to original shape/size (once force removed)Plastic: does not return to original shape/size (once force removed)

Tough: can withstand dynamic loads / shocks / impacts / absorbsa lot of energy before breaking2

(ii)Brittle:Snaps / cracks / shatters / breaks without (plastic) deformation(when subjected to a force)1

(iii)Strong:Large force / stress required to break it1

(b)(i)Breaking stress:Use of = ECorrect answer [2 108 Pa]

Eg. = 2 1011 0.001= 2 108 Pa2

(ii)Force to break wire:Use of A = r2Use of F = ACorrect answer [157 (N)][allow 156 157 (N) for rounding errors no u.e]

Eg.A = (1 103/2)2 = 7.9 107 m2F = 2 108 7.9 107 m2Weight (= F) = 157 N3

(iii)Force to break Biosteel fibre:3.1 103 N [allow 3.1 103 N 3.2 103 N]

eg.20 157 = 3140 N (3200 N if 160 N used)1

(iv)Assumption:Elastic limit (of both materials) not reached / elastic behaviour /Hookes law obeyed / Young modulus still holds at breaking point/ Area remains constant / best Biosteel scenario / 20 stronger1

[11]

120.(a)Angles:Normal correctly added to raindrop (by eye)

An angle of incidence correctly labelled between normal andincident ray and an angle of refraction correctly labelledbetween normal and refracted ray2

(b)Angle of refraction:Use of = sin i / sin rCorrect answer [20][allow 2021 to allow for rounding errors]

eg.sin r = sin 27/ 1.3r = 202

(c)(i)Critical angle:The angle beyond which total internal reflection (of the light)occurs [allow T.I.R] / r = 901

(ii)Critical angle calculation:

Use of = 1 / sin CCorrect answer [50.3] [allow 50 51]

Eg.Sin C = 1/1.3C = 50.32

(d)Diagram:i = 35 [allow 33 37]

Ray of light shown refracting away from normal on leavingraindropSome internal reflection of ray also shown with i = r [by eye]

Reflected ray shown refracting away from the normal as it leavesthe front of the raindrop / angle of refraction correctlycalculated at back surface4

(e)Refractive index:(Red light has) lower refractive index (than violet light)1[12]

121.(a)Graph:Line of best fit drawn as straight line through origin

Stiffness:Use of k = F/xk = 22 N m1 (allow 21 23 N m1)[allow ecf from graph]

eg.k = 110/5.0 = 22 N m13

(b)Energy stored in rope:Use of E = area under graph = Fx[also allow E = kx2 / E = F2/2k]Correct answer [514 J (511J if k = 22 Nm1 is used][allow 505 522 J]

E = 150 6.85= 514 J2

(c)Less extension with child:Any one of:Rope connected to sheet as (many) parallel sections

(making it stiffer)Each section of rope supporting (much) less than full weightWork done / energy lost to friction with trampoline framMax 1

(d)New rope dimensions:Rope needs to be thicker / shorterAs stiffness would have to be increased / reference toE = Fl/Ax / as it would have to withstand a greater stress /otherwise extension would be (much) greater2

[8]

122.(a)Show sum of quark charges in proton = +1+2/3 +2/3 1/3 = (+) 1 (1)

Show sum of quark charges in neutron = 0+2/3 1/3 1/3 = 0 (1)

[ignore references to e]2

(b)(i)baryon (1)meson (1)2

(ii)baryon: 3 quarks (1)meson: quark/antiquark (1)[1 for answers reversed or baryon/meson not specified]2

(c)any 4 marks from the following examples:high speed means high energy/momentum (1)may need to overcome (electrostatic) repulsion (1)more energy available for creating particles (1)higher energy/momentum/speed means shorter wavelength (1)reference to = h/mv or = h/p (1)for diffraction/scattering (1)need approx equal to particle spacing/internal structure (1)max 4

(d)Speeds near the speed of light (1)[11]

123.Recall speed = s/t (1)Use of s = D (1)Answer for speed (1)Conclusion (1)

ORUse of v = rUse of = 2 20 000Answer for speedConclusion

v = s/ts = 8000 (m)v = 8000 20 000 (m/s)v = 5 108 m/s

inaccurate/not possible since speed > c

[4]

124.(a)Lines (1)[not crossing; minimum 2 lines starting from S pole of magnet]Correct arrow(s) (1)[minimum 1 arrow pointing towards S pole, any incorrect arrow scores 0]2

(b)(i)Use of F = BIl rearranged to B = F/Il OR with two correct subs (1)Leading to correct answer (1)

B = F/Il = 0.008/(5.8 0.012) (T)B = 0.11 (T)2

(ii)Assumption:parallel field/uniform field/constant field for 12 mm then falls to zero /assume wire perpendicular to field (at all points)/ = 90 (where F=Bilsin given earlier)/force same at all points on the wire (1)1

(c)Experimental value less because field divergesOR field strength decreases with distanceOR field could be 0.3 T at magnet surface and only 0.1 T at wire (1)1

(d)Wire would levitate (again) (1)Two reversals cancel/applying FLHR (1)[wire moves downwards due to current OR field reversed scores 1]2

[8]

125.(a)capacitors need d.c. (1)OR Mains is a.c. / mains current changes direction constantly

Charge given in one half of cycle is removed the next half (1)OR C charged then discharges2

(b)(i)Voltage value for initial voltage 1/e [or use of 37%] (1)OR use 2 values where 2nd is 1/eth of firstOR draw tangent at time = 0 sOR V = Vo et/RC with correct substitution of (t,V) from graph

T = 0.07 s [allow 0.065 s 0.075 s] (1)2

(ii)Recall time constant = CR (1)Answer for R [allow ecf for T] (1)

R = T/C = 0.07 s/(100 109 F)R = 7 105 2

(c)(i)Recall Q = CV [equation or substitution] (1)Answer for Q (1)

Q = CV = 100 106 300Q = 0.03 C2

(ii)Recall W = 1/2 CV2 OR W = 1/2 Q2/C (1)OR 2 correct subs into W = 1/2 QV [allow ecf]Answer (1)

Eg: W = 1/2 QV = 0.5 0.03 300 (J) = 4.5 JOR W = 1/2 CV2 = 0.5 100 109 300 300 (J) = 4.5 JOR W = 1/2 Q2/C = 0.5 0.03 0.03/(100 109) (J) = 4.5 J2[10]

126.Any 8 marks from:Recall of p = mv (1)Use of momentum before collision = momentum after collision (1)Correct value for speed (1)

Example:1250 28.0 + 3500 x25.5 = (1250 + 3500) vv = 26.16 m s1

Recall of ke = 1/2 mv2 or ke = p2/2m (1)Total ke before (1)Total ke after (1)Loss in ke (1)Recall of work = force distance (1)Correct answer for force to 2 SF (1)

Example:Total ke before = 1 138 000 J + 490 000 J = 1 627 938 JTotal ke after = 1 625 059 JLoss of KE = 2879 JBraking force = 2879/5 = 576 NMax 8

[8]

127.(a)Name of effectResonanceIdea that step frequency = natural frequency of bridge2

(b)(i)Why oscillations are forcedE.g. amplitude is increasing OR oscillations are driven by the

wind1

(ii)Calculation of maximum accelerationUse of = 2 / T to obtain value for Correct answer for acceleration [14 m s2]

Example of calculation: = 2 / T = 2 /(60 s / 38) = 4.0 s1amax = 2 A = (4.0 s1)2 0.90 m = 14 m s22

(iii)Show that car loses contactRequired amax is greater than gSo, at a position (of 0.61 m) above the equilibrium

position, vehicle loses contact with the road

Example of calculation:x = g/ 2 = 9.81 m s2/(4 s1)2 = 0.61 m2[7]

128.(a)Name of reaction(Nuclear) fusion1

(b)(i)How to determine distance of starExplanation using F = L / 4 d2 (Use known luminosity with

measured flux at Earth to determine d)1

(ii)How to determine velocity of starMention of Doppler Effect OR red shiftIdentify (pattern of) lines and compare with lab

frequencyf (relative) v OR the greater the velocity of the star

(relative to Earth), the greater the change in

frequency/wavelength observedMax 2

(c)(i)Line of best fitInsertion of line of best fit, through origin 1 square, with

approx. the same number of points each side of lineIdea that the greater the distance to the galaxy, the

greater its velocity (relative to Earth)2

(ii)Use of gradient to calculate age of universeUse of v = Ho d to argue that Ho is the gradient of the graphCorrect answer for age of universe[4.6 1017 s, accept 4.0 1017 s 5.2 1017 s]

Example of calculation:1/gradient = 120 106 pc 3.09 1016 m pc1/ 8,000 103 m s1 =4.6 1017 s2

[8]

129.(a)(i)Meaning of symbolsm = mass of a gas molecule = mean square speed of gas moleculeT = absolute temperature [accept kelvin temperature]3

(ii)Physical quantity represented(mean) kinetic energy (of a gas molecule)1

(iii)Calculation of velocityUse of m< c2 > = 3/2 k T with T = 223 KCorrect answer for velocity [410 m s1]

Example of calculation:c = (3 1.38 1023 J K1 223 K / 5.4 1026 kg) = 413 m s12

(b)(i)Obtain expression for escape velocityIdea that total energy must be zero for molecule just to

escapeSo, m vesc2 GMm/r = 0, leading to required equation2

(ii)Show that escape velocity > 10 km s1Use of vesc = (2GM/r) with r = (6.37 + 0.01) 106 mCorrect answer for escape velocity [11.1 km s1, at least 2

sig. figs. required]

Example of calculation:vesc= (2GM/r)vesc= (2 6.67 1011 N m2 kg2 5.98 1024 kg /(6.37 + 0.10) 106 m)

= 1.11 104 (m s1)

= 11.1 (km s1)2

(iii)Use of graph to explain whether molecules are likely to escapeIdea that only a tiny fraction of molecules have a very

high velocityAny quantitative attempt to compare the r.m.s. velocity

with the escape velocity leading to the conclusion that

molecules are not likely to escape. e.g. 410 is much less

than 11,0002

[12]

130.(a)Change in nuclear compositionNucleus has one less neutron OR nucleus has one more proton)1

(b)(i)Calculation of age of skullUse of = ln2/t to obtain value for Use of N = NoetCorrect answer for age of skull [1.2 104 y; 3.83 1011 s]

Example of calculation: = ln 2/t = ln 2/5730 y = 1.2 104 y1 [3.84 1012 s1]ln(N/No) = tln(2.3 1011/1.0 1010) = (1.2 104 y1)tt = 1.2 104 y

Alternative mark schemeUse of half life ruleCorrect answer for number of half lives [2.12]Correct answer for age of skull [1.2 104 y]

Example of calculation:N/ No = (0.5)n(2.3 1011)/(1 1010) = (0.5)nlog(0.23) = n log(0.5)n = log(0.23)/log(0.5) = 2.12t = 2.12 5730 = 1.2 104 y3

(ii)Reason for inaccuracyIdea that it is impossible to know the exact proportion of

14C in the atmosphere when the bones were formed OR

reference to the difficulty of measuring such small

percentages of 14C.1

(iii)Why 210Pb is more suitable:Idea that the half life of 210Pb is closer to the age of

recent bones [e.g. a greater proportion of 210Pb will have

decayed as the time elapsed is one or more half lives]1

[6]

131.(a)181181 (1) O +p/H equals F+ n (1)8190 (1)[omitting the n with everything else correct = 1]

3(b)Accelerated through 19 106 V / MVUsing linear accelerator / cyclotron / particle accelerator / (1)recognisable description (1)2

(c)Time taken for half the original quantity/ nuclei /activity to decay (1)

Long enough for (cancer/tumour/body to absorb) and still beactive/detected (1)

Will not be in body for too long (1)3

(d)Use of E = mc2 (1)Use of E = hf (1)Use of v = f (1) = 2.4 1012 m (1)

eg 9.11 1031 9 1016 (2)f = 8.2 1014 / 6.6 1034 ecf = 3 108 / 1.2 1020 ecf4

(e)Conservation of momentum (1)

Before momentum = 0 (1)

so + for one photon and for other (1)2 max

[14]

132.(a)Show that the speed is approximately 30 m s1Sets EK = mgh (1)Substitution into formulae of 9.8(1) m s2 or 10 m s2 and 50 m. (1)[Also allow substitution of 60 m for this mark]Answer [31 m s1. 2 sig fig required. No ue.] (1)Egmv2 = mghv2 = 2gh = 2 9.81 m s2 50 mv = 31.3 (m s1) Answer is 31.6 m s1 if 10 m s2 is used

Also allow the following solution although this is notuniformly accelerated motion.

v2 = u2 + 2asv2 = 0 + 2 9.81 m s2 50mv = 31.3 (m s1)3

(b)(i)Average braking force [ecf value of vlFor the equation mv2 (1)[give this mark if this is shown in symbols, words or values]Attempts to obtain the difference between two energyvalues that relate to with and without the braking system orfor setting an energy value equal toForce 80 m (1)Answer [800 N if 30 m s1 used; If 31.3 m s1 or 31.6 m s1 areused accept answers in the range 1100 N 1300 N; If 34 m s1is used answer is 2000 N] (1)

Eg (367875 J)ke after free fall (273375 J)ke at 27m/s = 94500 JF 80 m = 94500 JF = 1180 N

Also allow the following solution.

Selects v2 = u2 + 2as and F = ma (1)Attempts to obtain the difference between two forces /accelerations that relate to with and without the brakingsystem. (1)Answer [800 N if 30 m s1 used; If 31.3 m s1 or 31.6 m s1 areused accept answers in the range 1100 N 1300 N; If 34 m s1is used answer is 2000 N] (1)

Eg Braking force = () 750()= () 1175 N3

(ii)Why braking force of this magnitude not requiredAir resistance (would also act to reduce speed) (1)Or Number and/or mass of passengers will varyOr Friction [ignore references to where forces act for thismark. A bald answer ie friction is acceptable]Or Accept some (kinetic) energy is transferred [not lost] tothermal energy [accept heat] (and sound)Or Work is done against friction1

(iii)Explain whether braking force would changeQWOC: (1)EitherThe kinetic energy will be greater (because the mass of thepassengers has increased) (1)(hence) more work would have to be done(by the brakingsystem) (1)(The distance travelled, P to Q, is the same therefore)greater (braking) force is required (1)

OrMomentum (of the truck) will be greater (because the massof the passengers has increased) (1)Rate of change of momentum will be greater or [allow] thetime taken to travel (80 m) will be the same [if thecandidate writes constant allow this if you feel they meansame] (1)(Therefore) greater (braking) force is required (1)

Or(Allow) Change in velocity and the time taken (for the truckto travel 80 m) will be the same or (Average) deceleration /acceleration will be the same [accept constant if theymean same. Also accept any fixed value for accelerationeg 9.8 m s2] (for greater mass of passengers) (1)(since) F = ma and mass has increased (1)A greater (braking) force is required (1)4

[11]

133.(a)(i)Additional heightAnswer [ 5 (m)] (1)

Eg distance = area of small triangle = 0.5 1 s 10 m s1 = 5 m1

(ii)Total distance travelled [Allow ecf of their value]Distance travelled between 1 s and 4s [45 m] (1)Answer [ 50 m] (1)

Eg distance fallen = area of large triangle

= 0.5 3 s 30 m s1

= 45 m total distance =45m + 5m = 50m2

(b)Objects displacement40 m (1)Below (point of release) or minus sign (1)[Ecf candidates answers for additional height and distance ieuse their distance 2 their additional height]2

(c)Acceleration time graphLine drawn parallel to time axis extending from t = 0 (1)[Above or below the time axis]The line drawn parallel to the time axis extends from 0 s to 4 s (1)[If line continues beyond or stops short of 4 s do not give this mark]Acceleration shown as minus 10 m s2 (1)[This mark is consequent on the second mark being obtained]3

[8]

134.(a)Account for the forceWhen the flea pushes (down) on the surface the surface[accept ground, not earth] pushes back / upwards (1)with an equal (magnitude of) force (1)[A statement of Newtons 3rd law gets no marks it must be applied]2

(b)(i)Show acceleration is about 1000 m s2Either Selects v2 = u2 + 2as Or two appropriate equations ofmotion (1)Correct substitution into the equation (1)[Do not penalise power of ten error. Allow 0.4 mm and 0.9 ms1 substitutions for this mark.]Answer [in range (1025 1060) m s2, must be given to atleast 3 sig fig. No ue] (1)

Eg (0.95 m s1)2 = 2 a 0.44 ( 103) ma = 1026(ms2)

OrSets changing Ke = work done (as legs expand) (1)Correct substitution into the equation (1)Answer [1030 m s2, must be given to at least 3 sig fig. No ue] (1)

Eg Ke = average F height m (0.95 m s1)2 = m a 0.44 (103) ma = 1026 (m s2)3

(ii)Resultant force(Allow ecf)Answer [4.1 104 N. 4(.0 ) 104 N if 1000 m s2 used. Ue.] (1)

Eg Force = 4 107 kg 1030 m s2 = 4.12 104 N1

(c)(i)What constant force opposes upward motionThe weight of / gravitational attraction / gravitational force /gravitational pull / force of gravity / accept pull of earth(on flea) (1)[Not just gravity. Accept bald answers ie weight]1

(ii)Change in heightSelects s = ()t or uses v = u + at (to find a) then eitherv2 = u2 + 2as or s = ut + at2 (1)Correct substitution (1)[If two equations are used a is negative]Answer [4.4(2) cm. Do not accept 4.5 cm] (1)[Nb the correct answer can be obtained from omitting ut andusing +a this would get 1 /3]

[Use of s = ut + at2 or v2 = u2 + 2as with lal = g and u = 0.95m s1 will get 1/3 if no attempt is made to find a. Forcandidates who use a = 1000 m s2 from (b)(i) give no marks]

Eg s = 9.3 102 s= 0.0442 m

Ora = s = 0.95 m s1 9.3 102 s + 10.2 m s2 (9.3 102 s)2= 0.0442 mor 0 = (0.95 m s1)2 + 2 10.2 m s2 s hence s = 0.0442m3

[10]

135.(a)Identify particleAlpha (particle) / Helium nucleus/////1

(b)Momentum of particleMomentum equation [In symbols or with numbers] (1)

EitherCorrect substitution into mv2 = energy (1)Use the relationship to determine the mass [6.6 1027 kg] (1)Answer [9.3 1020 (kg m s1) Must be given to 2 sig fig. Nounit error] (1)

OrRearrangement of Ek = mv2 to give momentum ie (1)Correct substitution (1)Answer [9.3 1020 kg m s1. Must be given to 2 sig fig. Nounit error] (1)

Eg m(1.41 107 m s1)2 = 6.58 1013 Jm = = 6.6 1027 kgmomentum= 6.6 1027 kg 1.41 107 m s1

= 9.3 1020 (kg m s1)

OrMomentum = = 9.3 10 (kg m s1)4

(c)Consistent with the principle of conservation of momentum(Since total) momentum before and after (decay) = 0 (1)State or show momentum / velocity are in opposite directions (1)[Values of momentum or velocity shown with opposite signswould get this mark]

Calculation ie 3.89 1025 kg 2.4 105 m s1 = 9(.3) 1020 (kg m s1) (1)

Eg 3.89 1025 kg 2.4 105 m s1 = 9(.3) 1020 kg m s13

[8]

136.(a)Calculate the ratio the densities of the atom and the nucleusDensity equation [In symbols or numbers] (1)Show the relationship between density and radius. (1)[Candidates who start by stating that density is inverselyproportional to the radius cubed would get both these marks.Candidates who show an expression where the mass isdivided by would set both these marks. Candidates whowrite Ratio = (1/105)3 would get both of these marks.]Factor 1015 established. [Some working must be shown forthis mark] (1)

Eg (Density)atom = or Density (Density)nucleus =

= (105)3

Assumption (entire) mass of the atom is concentrated in thenucleus[there must be a reference to the nucleus] (1)[eg mass of the atom =/approx mass of the nucleus; most /majority of the atoms mass is in the nucleus. The followingwould not be awarded marks; The atom is mostly emptyspace; mass of the electrons is negligible; the nucleus is avery dense.]4

(b)ObservationA very small percentage of particles [accept very few notjust a few. Do not accept some] are deflected throughangles greater than 90 / are back-scattered / deflected back. (1)[Allow; nearly all / most (alpha) particles pass through(the atom) without being deflected (showing the atom isvirtually empty space).][Accept nearly all, not many for the word most.]1

[5]

137.(a)How a beta-minus particle ionisesWhen a beta particle removes [accept repel] an electronfrom an atom / molecule (1)1

(b)How ionisation determines rangeState that each ionisation requires energy (1)The energy (to ionise) is obtained from the (transfer of)(kinetic) energy of the beta particle (which is therefore reduced) (1)Along its path it produces many ionisations until all its(kinetic) energy is used up (1)The more ionising a particle the shorter its range or the lessionising the greater the range (1)[Candidates may give the wrong reason for ionisation or evencompare alpha and beta but still award this mark.]Max 3 marks from 4[Note that the word kinetic is not essential for marks 2 and 3]3

(c)Why more ionisation is produced towards the end of its range(Towards the end of its range) the beta particle is travellingslower or has less kinetic energy (than at the beginning of its range) (1)(as a result it takes longer travelling a given length) andtherefore has more (close) encounters with atoms / moleculesor more opportunities to ionise (atoms / molecules)or will remain in contact (with atoms / molecules) longeror will collide with more (atoms / molecules per unit length)or ionisation (of atoms/molecules) is more frequent (towards end of range) (1)2

[6]

138.(a)n is (number of) charge carriers per unit volume ornumber density or (number of) charge carriers m3 orcharge carrier density(1)

[allow electrons]

v is drift speed or average velocity or drift velocity(of the charge carriers) (1)

[just speed or velocity scores zero]2

(b)/ A and Q A s or / Cs1 and Q C (1)n m3 (1)A m2 and v m s1 (1)[If no equation written assume order is that of equation]3

(c)(i)n 1 and Q Need all three1

(i)Ratio vA/ vB less than 1 following sensible calculation (1)Ratio = // 0.25 // 1:4 (1)(ratio 4:1 scores 1)[4vA:1vB scores 1]2

[8]

139.(a)Use of P = IV (1)Current in lamp A 2 A (1)

[0.5 A scores zero unless 24 = I 12 seen for 1st mark]2

Example of answerI = P V = 24 W 12VI = 2A

(b)(i)Voltmeter reading = 12 V (1)1

(ii)p.d. across R2 = 6 V or their (b)(i) minus 6V (1)Use of R = V/I (1) conditional on first markR2

Answer to this part must be consistent withvoltmeter reading and if voltmeter reading is wrongthis part has a max 2. If (b)(i) = 15 V then need to see

If (b)(i) = 6V or less they are going to score zero for this section.3

(iii)current through R1 = 5 A (1) ecf answers from (a)1

Example of answerCurrent through R1 = 2 A + 3 A = 5 A

(iv)p.d. across R1 = 3 V (1) ecf (15V minus their (b)(i))1

Example of answerp.d. across R1 = 15 V 12 V = 3 V

(v)R11

Example of answerR1 = 3 V 5A = 0.6[accept fraction 3/5]

[9]

140.(a)(i)EI (1)1

(ii)I2R (1)1

(iii)I2r (1)1

(b)EI = I2R + I2r or E = IR + Irecf Must use values (a)(i)-(iii)1

(c)I for circuit given by Imax = E / r or substitution of5000V into the equation (1)(for safety) need I to be as small as possible (1)3

[7]

141.(a)(i)Reference to a temperature related gas law (1)[V/T= constant or p/T= constant or pV/T = constantor pV = nRT; just symbols acceptable or word equivalentbut not Pressure law or Charles law]At absolute zero, V = zero or p = zero or pV = zero (1)2

(ii)temperature or at absolute zero the molecules haveno kinetic energy (1)[do not accept depends on, is related to etc]

(implies) at absolute zero molecules stationary ornot moving or still or speed / rms of molecules is zero (1)[If particles/atoms used for both statements 1/2]2

(b)Kelvin [absolute thermodynamic scale]1

(c)(i)Use of pV = nRT (1)Use of 300 K (1)n = 4.0 moles (1)mass of air = 0.12 kg (1) ecf their n[If no temp conversion n = 45 moles, mass = 1.3 kg scores 2/4]4

Example of answern = (1.0 105 Pa 0.10 m3) (8.31 J K1 mol1 300 K)n = 4.0 molesmass of air = 4.0 moles 0.029 mole kg1 = 0.12 kg

(ii)nT = constant oror calculation ofinitial density (1)correct use of above equations using Kelvintemperatures or calculation of final density (1)ratio = 3/5 or 0.6 (1) consequent on gaining method marks

[ratio of 1.7 or 5/3 could score 2][calculation of new mass in oven = 0.072 kg scores 1]

[If Kelvin not used in (c)(i) do not penalise here.Ratio is 0.12 again consequent on method marks]3

Example of answer227 C = 500K 27 C = 300 K 500 300 = 300 500 = 0.6

[12]

142.(a)(i)Volume of gas (1)amount of gas or mass of gas or number of moles of gas (1)2

(ii)Suitable diagram to include following labelled items(Trapped) mass of gas (1)method of indirectly heating gas (1)pressure gauge/reader/scale/mercury manometer (1)thermometer (1)

[wrong experiment e.g. Boyles Law 0/4]4

(iii)precaution;Minimise amount of gas not in water bath, stirring,allowing time for gas to reachtemp, parallax errors,ANY ONE

[not insulating the beaker or the water bath][not repeat readings]1

(b)Axes labelled with variables and units (1)Straight line graph with positive gradient (1)+ve intercept on pressure axis and meeting temp axisat 273 C OR graph through origin if Kelvin scaleused and zero written where axes cross. (1)

[if variables other than p and T used 0/3]3

[10]

143.(i)Sun now & as red giant

Similarity: (nuclear) fusion / burning (in core) OR mass (1)

Difference: H vs. He fusion / r.g. lower (surface) T/ r.g. higher core T /r.g. lower (mean) density [assume r.g. referred to if not specified] (1)

(ii)White Dwarf star terms

Hot, low, (surface) area, off (the main sequence)

Any three correct (1)

All four correct (1)4

[4]

144.(i)Large mass star fusion rate

Quality of written communication (1)

(30 M star) fuses at a greater rate AND spends less time on m.s. (1)[accept power, luminosity]

(30 M star) has greater temperature / (gravitational) forces /pressure (1)

leaves main sequence after hydrogen (and/or He) burning ceases /H fuel depleted (in core) [accept H used up] (1)4

(ii)2.2 M core remnant

Neutron star (1)1

(iii)Sun evolutionary phases

Red giant (1)

White dwarf (1)

Black dwarf (1)

[1 mark per error only if more than three phases circled]3

[8]

145.(i)Stress-strain graph regions

Neckings C or D (1)

Elastic deformation = A or B (1)

Plastic flow = E [ignore extra C or D] (1)3

(ii)Young modulus calculation

Attempt at gradient / stress strain (1)

Sensible pair of values [from linear region, ignore 10n] (1)

1.35 [allow 1.30 1.40] (1)

1011 Pa [or N m2] (1)4

(iii)Second material

Straight line [allow slight curvature at end] (1)

Less steep than original line (1)

Stops at = 2.6 103 (1)3

[10]

146.(a)Calculation of angular speed

Use of = 2/T (1)7.27 105 [2 sig fig minimum] (1)2

2/(24h 3600 s h1) = 7.27 105 rad s1(b)(i)Calculation of accelerationUse of a = r2 OR v = r and a = v2/r (1)0.034/031 m s2 (1)2

(6400 10J m)(7.27 105 rad s1)2= 0.034 m s2(ii)Direction of accelerationArrow to the left (1)[No label needed on arrow. If more than one arrow shown, nomark unless correct arrow labelled acceleration]1

(iii)Free-body diagramArrow to left labelled Weight/W/mg/pull of Earth/gravitationalforce (1)Arrow to right labelled Normal reaction/N/R/push of Earth (ORground)/(normal)contact force (1)

[Dont accept gravity as label][More than two forces max 1][Diagram correct except rotated gets 1 out of 2]2

(iv)How the acceleration is producedN is less than W (1)Resultant (OR net OR unbalanced) force towards centre (1)

[Accept downward / centripetal for towards the centre, butnot as an alternative to resultant]2

[9]

147.(a)Experiment

Scheme for timing methods:

QOWC (1)Use f = 1/T OR f = (number of cycles/time taken) (1)Apparatus (1)Principle of method (1)One precaution for accuracy (1)Max 4

Examples for last three marks:

Stopclock / stopwatch (1)Measure time taken for a number of cycles (1)

ensure vertical oscillations / not exceeding elastic limit (1)OrMotion (OR position) sensor and datalogger (OR computer) (1)Read time for one (or more) cycles from displacement-time graph (1)Read time for several cycles / ensure vertical oscillations (1)OrLight gate and datalogger (OR computer) (1)Computer measures time interval between beam interruptions (1)Use narrow light beam / position gate so beam cut at equilibriumposition / ensure vertical oscillations (1)OrVideo Camera (1)Read time for one (or more) cycles from video (1)Read time for several cycles / ensure vertical oscillations (1)

[Mark other reasonable techniques on the same principles]

Scheme for strobe method:

QOWC (1)Use stroboscope [Accept strobe] (1)Adjust frequency until mass appears at rest (1)Find highest frequency at which this happens (1)Repeat and average / ensure vertical oscillations (1)Max 4

Scheme for use of T m/k)/ T e/gQOWC (1)Calculate T from measured (OR known) m and k usingTm/k) (1) or calculate T from measured e known g usingTe/g (1)Use f = 1/T (1)Max 2[e is the extension of the spring produced by weight of mass m]

[Do not give any credit for experiments to measure the resonantfrequency of the system]

(b)(i)GraphAxis labels and single peak (1)Rounded top and concave sides (1)f0 marked on the frequency axis at, or just to right of, peak.[Amplitude at f0 should be at least 75% of maximum amplitude] (1)

[A sharp kink loses mark 2 only][Graphs with multiple peaks lose marks 1 and 2; f0markedcorrectly on lowest frequency peak for mark 3][Ignore whether or not curve goes to origin]3

(ii)Name of phenomenonResonance (1)

[Mark this independent of whether graph is correctDo not accept resonant frequency]1

(iii)Footbridge applicationPeople walking / wind / earthquake can cause vibration / act as adriver / apply regular impulses (1)If resonance occurs OR if frequency equals / is close to f0 we mayget large / dangerous / violent oscillations OR large energytransfer OR damage to bridge (1)2

[10]

148.(a)(i)How we know the speed is constant

Crest spacing constant / circular crestsOr wavelength constant / equal wavelength (1)

[Accept wavefront for crests][Dont accept wave]1

(ii)Calculation of speed is 10 mm (1)[Allow 9 to 11]Use of v = f (1)0.40 m s1 (1)

[Allow 0.36 to 0.44Allow last two marks for correct calculation from wrong wavelength]3

(40Hz)(10 103 m)= 0.40 m s1(b)Line X1st constructive interference line below PQ, labelled X (1)

[Accept straight lineIgnore other lines provided correct one is clearly labelled X]1

(c)(i)Superposition along PQConstructive interference / reinforcement / waves of largeramplitude / larger crests and troughs (1)Crests from S1 and S2 coincide / waves are in phase / zero phasedifference / zero path difference (1)Amplitude is the sum of the individual amplitudes (OR twice theamplitude of the separate waves) (1)3

(ii)TableA constructive (1)B destructive (1)2

[10]

149.(a)Amplitude(i)Amplitude remains constant (1)1

(ii)Amplitude decreases then increases (1)Amplitude is zero at node (OR half way between X and Y) (1)2

(b)Phase difference

(i)Phase difference increases / is proportional to distance XP (1)1

(ii)Up to node phase difference is zero / in phase (1)Beyond the node phase difference is / 180 / half a cycle / inantiphase (1)

[Do not allow completely out of phase]2

[6]

150.(a)Part of spectrumLight / Visible / red (1)1

Calculation of work functionUse of = hc/ (1)3.06 1019 [2 sig fig minimum] (1)2

(6.63 1034 J s)(3.00 108 m s1)/(6.5 107 m)= 3.06 1019 J

(b)(i)Meaning of stopping potentialMinimum potential difference between C and A / across thephotocell (1)Which reduces current to zero OR stops electrons reaching A /crossing the gap / crossing photocell (1)2

(ii)Why the graphs are parallelCorrect rearrangement giving Vs = hf/e /e (1)Gradient is h/e which is constant / same for each metal (1)

[Second mark can be awarded without the first if norearrangement is given, or if rearranged formula is wrong butdoes represent a linear graph with gradient h/e]2

[7]

| 151.(a)Calculation of recession speed = 684 656 (1)Use of v/c / (1)1.28 107 m s1 (1)

[Substituting 684 for , leading to 1.23 107, loses last two marks]3

(3.00 108 m s1)(28 109 m)/(656 109 m)= 1.28 107 m s1(b)Calculation of wavelength received from YBy Hubbles law / v = Hd / v proportional to d, as d is doubled,v is doubled (1) = 56 / is doubled (1)712 nm (1)

[Bald answer of 712 nm, with no working or explanation, gets 2marks only]

[If candidate gets part (a) wrong, accept EITHER 56, 712 for thelast two marks (if they have avoided reusing the formula or madethe same mistake again) OR ecf (if they have repeated thecalculation but avoided the original mistake)]3

[6]

152.(a)(i)P.d. across capacitor

Use of VR = I R (1)

[allow one error of 103 in individual substitutions; disallow ifVR value is 6V]

VC = 6.0 V 4.0 V (= 2.0V) (1)[No ecf]2

Example of answer:

VR = 20 106 A 2.0 105 = 4.0 V

Hence Vc = 6.0 V 4.0 V = 2.0 V

(b)Calculation of charge

Use of Q = C V with 560 F & 2.0 V (1)

[Check correct equation is being used; allow power of 10 error incapacitance value. If capacitance value mis-transcribed, allow thisfirst mark only]

Answer 1.1 (2)mC (1120C) [no ecf] (1)2

(c)Calculation of energy stored

Use of W = CV2 with given values, or W = V with their Q,to get 1.1(2) mJ (1120J) or their correct answer. (1)[same numerical value as in (b)]1

(d)Calculation of energy transferred

Use of E = QV, with their Q and V = 6.0 V, to get 6.7(2) mJ (6720J)or their answer [6 value at part c] correctly found. (1)1

(e)Main reason for energy difference

Energy is transferred to thermal / heat energy in / work is doneagainst, the resistance of the resistor in the circuit [NOT just the resistance of the wires, nor the components] (1)

[Do not credit vague reference to energy dissipation, nor energyis lost to the surroundings]1

[7]

153.(a)Calculation of potential difference

Use of mpv2 with v = 2.77 105 m s1

and mp = 1.67 1027 kg (1)

Use of eV with e = 1.60 1019 C (1)[beware confusion of v and V]

Answer = 400(.4) / 401 V (1)[If data used to 2 sf, 380V, 384V or 364V, allow 2/3]3

Example of answer:

eV = mpv2V = = 400V[beware unit error of eV here]

(b)Add second path to diagram

Path at B stays equidistant from that at A [gauge by eye] (1)1

(c)(i)Add path to diagram

Added path at A [allow through letter A] also curves upwards (1)

But is less curved than the original, straight beyond plates andcontinues to diverge from it (1)2

(ii)Explanation

Charge on a is double that on proton / has 2 protons /force on a is double force on proton. (1)

Mass of a particle is (approx) 4 times / more than double that of theproton. (1)

[hence acceleration is approximately halved].

[Ignore reference to F = Bqv; do not credit referenceto unless implication of numbers 4 and 2 is made clear].2

[8]

154.(a)Two deductions [not simplv word descriptions of features of the diagram]

The gravitational potential is increasing with height / when movingaway from the Earth / Work must be done to move away from the Earth (1)

[Ignore idea that V ; in words or symbols]

The field is non-uniform / radial / Field strength decreases withheight / when moving away from the Earth (1)2

(b)Entry speed at Earths atmosphere

| 1MJkg1 ( 61MJkg1) = 60MJkg1 [accept 60MJkg1] (1)

Loss of GPE/ Gain in KE of spacecraft = m (1)

Statement / use of mv2 = mV OR v2 = V/(= 60 MJ kg1) (1)[either of these statements also earns the second mark, if not already awarded][See v2 = 1.2 108

Answer 1.095 103 m s1 /10950 m s1/11.0 km s1 [more than 2 sf] (1)4

(c)Showing relative distance

Use of Newtons Law; FE = or FM = (1) (1)[or equivalent re-arrangement] [or equivalent] (1)[correct relationship, expressed in terms of numerical values][If inverted, then MM:ME = 0.0123]So (independent mark) (1)[Stating 81 =100 at the third mark stage does not, alone, earn the 4th mark]

[Beware ambiguity or transposition of r values at steps 2 or 3]4

[10]

155.(a)(i)Direction of current

Position 1 = Q to P / anticlockwise / to the left} (1)Position 3 = P to Q / clockwise / to the right}[both needed; arrows added to diagram may givecurrent directions at 1 & 3]

Position 2 = no current (1)2

(ii)Current calculationUse of , or = Blv, = 2 102 T 0.12 m 0.05 m s1 (1)

[ignore power of 10 errors in dimension and velocity values]

(Emf =) 1.2 104 V (1)

I = or I = seen or used (1)

Answer = 6.0 105 A or 60A [ecf their emf] (1)4

(b)Uniform acceleration?

QoWC (1)

Magnitude of current would be increasing as frame movesthrough position 1 (or position 3) (1)

Magnitude of current would be greater for position 3 than 1[Beware comparison of position 3 with position 2 here] (1)

Reference to increased rate of flux cutting / increased rate}of flux change / increased area swept out per second} (1)(Beware suggestion that B or flux density is changing)}

So induced emf is greater (1)

Current for position 2 is zero}[Do not credit equal and opposite}currents cancelling]} Both needed (1)

Since flux linkage is constant /}(net) rate of flux cutting is zero /}Emfs in PS and QR are equal and opposite}Max 4

[10]

156.(a)(i)Calculate maximum current

Recall of P = IV (1)

Correct answer [0.49 A] (1)

Example of calculation:P = IVI = 5.9 W / 12.0 V= 0.49 A2

(ii)Show that resistance is about 24

Recall of V = IR (1)

Correct answer to 3 s.f. [24.5 ] [no u.e.] (1)

Example of calculation:R = 12 V / 0.49 A= 24.5 2

(b)(i)Calculate current

Use of correct circuit resistance (1)

Correct answer [0.45 A] (1)

Example of calculation:I =V / R= 12 V (24.5 + 2 )= 0.45 A2

(ii)Calculate power

Recall of P = IV and V = IR (accept P = I2R) (1)or P =

Correct answer [5.0 W] (1)

Example of calculation:P = I2R= (0.45 A)2 24.5 = 5.0 W2

(c)Increase in power available to pump

e.g. lower resistance in wire thicker wire, panel nearer to motor (1)(accept relevant answers relating to panels, e.g. more panels)1

[9]

157.(a)(i)Explain upward force is about 0.1 N

Correct answer for force to 2 s.f. [()0.092 N] [no ue] (1)

Explanation that negative means upwards (1)

Example of calculation:W = mg= 0.0094 kg 9.81 N kg1= 0.092 N2

(ii)Label balloon diagram ands show that weight is about 0.07 N

Tension + arrow (1)Weight + arrow (1)Weight = 0.068 N (1)

(Do not accept gravity for weight)3

(b)(i)Label 2nd balloon diagram

Weight (1)Air resistance (1)2

(ii)Expression for vertical component

T cos 43 / upthrust weight / 0.16 N 0.068 N / (1)(accept T sin 47)1

(iii)Calculate tension in string

Correct expression showing vertical forces on balloon (1)

Correct answer (0.13 N) (1)

Example of calculation:T cos 43 = 0.16 N 0.068 NT cos 43 = 0.092 NT = 0.13 N2

(c)Explain change in angle

Air resistance increases (1)

Horizontal component of tension increases (while verticalcomponent stays the same) (1)2

[12]

158.(a)(i)Name process

Refraction (1)1

(ii)Explanation of refraction taking place

change in speed / density / wavelength (1)1

(b)(i)Draw ray from butterfly to fish

refraction shown (1)

refraction correct (1)2

(ii)Explain what is meant by critical angle

Identify the angle as that in the denser medium (1)

Indicate that this is max angle for refraction OR total internalreflection occurs beyond this (1)[angles may be described in terms of relevant media]2

(iii)Explain two paths for rays from fish A to fish B

direct path because no change of medium/refractive index/density (1)

(total internal) reflection along other path /angle of incidence > critical angle (1)

direct ray correctly drawn with arrow (1)

total internal reflection path correctly drawn with arrow (1)

[lack of ruler not penalised directly] [arrow penalised once only]4

[10]

159.(a)Show that Ep lost is about 37 000 J

Recall of Ep = mgh (1)

Correct answer to 3 s.f. [37 300 J] [no ue] (1)

Example of calculation:Ep = mghEp = 760 kg 9.81 N kg1 5 m= 37278 J2

(b)(i)Show that Ek of projectile and counterweight is about 26 000 J

Correct calculation of Ep gained by projectile [10 800 J] [no ue] (1)

Correct calculation of Ek to 3 s.f. [26 200 J] [no ue] (1)

Example of calculation:Ep gained by projectile = 55 kg 9.81 N kg1 20 m = 10 800 JEk = 37 000 J 10 800 J= 26 200 J2

(ii)State assumption

All lost g