1 1 1 n 0) tt] 1 a iou. 2 14. 0) Wavelength =f2 M VT) (0.002 s] (1) [ive full credit for candidates who do this in | stage T= use of y Use of f: (f= 330/066 f= 0.66 /330 T= 0.0025 Direction of travel of light is water air 1) [Angle of incidence is greater than the critical angle (1) (a) Trmsverse waves osvllate in any direction perpendicular to wave direction (1) Longitudinal waves oscillate in one direction only OR parallel to way direction. (1) Polarisation reduces wave intensity by imiting oscillations and wave direction to only one plane OR limiting oscillations to one direction only. (1) (accept vibrations and answers in terms of an example such as a rope passing through sits) (b) Light source, 2 pieces of polaroid and detector e.g. eye, screen, LED OR laser, 1 polaroid and detector (1) Rotate one polaroid (1) Intensity of light varies () Frequency unaltered (1) ‘Wavelength decreases (I) Speed decreases (I) (a) Use of sensor Event happens very quickly OR cannot take readings fast enough (1) ‘Sampling rate: 50 samples per second (1) iO} (6(>) Initially the temperature is low so current is high Resistance of filament increases as temperature increases Current falls to steady value when temperature is constant Maximum heating is when lamp is switched on / when current is highest Filament breaks due to meting eaused by temperature rise Max 4 @ 1S. The answer must be clear and the answer must be organised ina logical sequence (QWC + Iwas known that X penewsated (1) + Tewasuot known hat X rays were harmafil (1) . Doctors died because of too much exposure (1), + Lack of shielding (1) 7 New treatments may have unknown side effects (1), . ‘Treatments need to be tested /time allowed for side effects to appear (1) Max 4 io 16 (@ [10m)a) 1 (b) Ratio of (5 or 6/3) « 60 (1) Answer [f= 100 Hz] (1) 2 fo + -New treatments may have unknown side effects (1) + Treatments need to be tested / time allowed for side effects to appear(1) Max 4 io}7 18, 19. 20. Useofsini/sinr=y (1) Use of either 80° or 1.33 (1) [r=48] 0) 3 Example of answer sin 80 /sinr= 133 [r= 48°] Both rays reftacted towards the normal ‘Violet refracted more than red. 2 [oT (a) Calculation of energy required by atom (1) Answer [1.8 eV) () Energy gained by stom = 13.6 eV-3.4 eV = 10.2 eV KE of electron after collision= 12 eV — 10.2 eV =1.8 eV 2 (b) Use of E=hfandc=/ (I) Conversion of eV to Joules (1) Answer=[1.22% 1077 m] (1) Example of answer E=hfand A a hcl (6.63 « 10 Js <3 x 10° ms) = (10.2eV * 1.6% 10 C) 21% 107m 3 ‘The answer must be clear, use an appropriate style and be organised in a logical sequence. (QWC) Reference to! = nqvd (I) For thelamp Increased atomic vibrations reduce the movement of electrons (1) Resistance of lamp increases with temperature (1) For the thermistor Increased atomic vibrations again reduce movement of electrons (1) ‘But increase in temperature leads toa large increase in n (1) ‘Overall the resistance of the thermistor decreased with increase in temperature, (1) Max 5 8) (a) Diffraction is the change in direction of wave or shape or wavefront (1) when the wave passes an obstacle or gap (1) 2aL (b) (©) Enetgy of photon increases as frequency increases OR reference to E=)f() Electrons require a certain amount of energy to break free and this corresponds to a minimum frequency (1) 2 (6) (8) (Use of speed =distance overtime (1) Distance =4 em ()_ Answer = [2.7 * 10° s}(0) Example of answer t=4em-= 1500 ms Ws 3 i) 2 (ii) Time forpulse to return greater than pulse interval (I) Allreflections need to reach transducer before next pulse sent. (1) ‘Will resat in an inaccurate image. (1) (Max 2) ‘Need to decrease the frequency of the ultrasound. (1) (Max 3) Max 3 (iv) X-mys damage cellsitissne'foetus/baby but ultrasound does not (need reference to both X-rays and ultrasound) (1) 1 The answer nustbe clear, use an appropriate style and be organised in a logical sequence (QWC) Doppler shiftis the change in frequency of a wave when the source or the receiver 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) 4 113] (a) Voltmeter is across resistor should be actoss cell (1) 1 (©) Plotof graph Check any three points (award mark if these are correct) (3) Line of best fit 3 Gi) emf =[1.36-144 V] 1Git) Attempt to find gradiem (1) Answer (0.38 -042 2] (1) 2 (6) Intercept would twice value above (1) (accept numerical value 2» value oo) Gradient would be twive value above (1) (accept numerical value 2% value Cyc), 23. (a) Diode or LED @) 1 (©) (Use of R= V / Teurrent borween 75 and 90 ignoring powers of 10 (1) © 24. answer 6780.0 (1) Example of answer 60 V=(85« 10) 4 06 2 i) Infinite OR very high OR 2 1 ANY ONE Rectification / AC to DC / DC supply [not DC appliances] Preventing earth leakage Stabilising power output ‘To protect components A named use of LED if linked to LED as component in (ay(eg. caleulator display / torch) A voltage controlled switch (Allow current in only one direction) 1 (5) (Resist - Resistivity = resistance » (1) cross sectional area / length (1) ‘p= Rall with symbols defined scores 2/2 equation as above without symbols defined scores 4 equation given as R= pl/d with symbols defined scores 1/2 (st mark is for linking resistivity to resistance with some other terms) 2(>) (Resistance calculation Converts KW to W (1) Use of P= V2/R OR P= Viand V=IR (1) Resistance = 530 (1) Example of answer Gi) Length calculation Recall R= pl/A (1) Correct substitution of values (1) Length = 6.3 m (accept 6.2 m) (1) ecf value of R Example of answer 1= (52.90 «1.3 «107 m?) + (Lx 1070 m) 1=63m 3 (ii) Proportion method Identifies a smaller diameter is needed (1) Diameter = 0.29 mm (1) OR Calculation method Use of formula with 1 = half their value in (b)(i) (1) Diameter = 0.29 mm (1) (Eef.a wrong formula from part ii for full credit) Example of answer Aggy = OAT mm + V2 hey =0.29 mm, 2 [10 B’25.(a) Definition of EMF. Energy (conversion) or work done (1) Per unit charge (1) [work done/coulomb 1/2, energy given to a charge 1/2, energy given to a charge of a coulomb 2/2] OR OR E=WQd) E=Pt Symbols defined (1) Symbols defined (E=1 VC scores 1) (E= 1 WIA scores 1) (Terminal) potential difference when no current is drawn 1/2) 2(b) (ii) © @ (@ Internal resistance caleulation Attempt to find current (1) Pd across r= 0.2 V (1) = 0.36) (1) [You must follow through the working, Thave seen incorrect methods getting 0.36 2] Example of answer 8V +500 3.0-2.8) V+ 0.56 A= 0362 3 Combined resistance ‘Use of parallel resistor formula (1) Resistance = 3.3 Q [accept 31/3 but not 10/3] (1) 2 Voltmeter reading (ecf bil) Current calculation using 3 V with either 3.3 Q or 3.7 Q (1) Total resistance = 3.7 Q [accept 3.66 10 3.73 Q] OR use of = E— Jr (1) Voltmeter reading = 2.7 V (1) OR Potential divider method, ratio of resistors with 3.7 0 on bottom (1) Multiplied by 3.0 V (1) 27VQ) Example of answer Veotmeer= 3.32% 081A =2.7V 3 Ideal voltmeter {deal voltmeter has infinite resistance OR extremely high resistance OR highest possible R OR much larger resistance than that of ‘component it s connected across OR quotes value > I MQ (1) ‘Current through voltmeter is zero (negligible) OR doesn't reduce the resistance of the cireuit OR doesn’t reduce the pd. itis meant to bbe measuring. (1) 2 (ny ai Potentiometer correctly connected ie potential divider circuit (1) Ammeter in series and voltmeter in parallel with bulb (1) (ight bulb in series with resistance can score sevond mark ony)) nL @ (i) (a) Graph “+1, FV quadrant; curve through origin with decreasing gradient (1) [do not give this mark if curve becomes flat and then starts going down ie. ithas ahook] “I, -V quadrant reasonably accurate rotation of +1+V quadrant (1) [As current voltage increases, temperature ofthe lamp increases / lamp heats up (1) Leading to increase in resistance of lamp (1) Rate of increase in current decreases OR equal increases in V lead to smaller increases in I (1) Qowe (1) Ecfifa straight lie graph is drawn max 3 R constant (1) Vall) Qowe (1) (i) Demonstrating the stationary wave ‘Move microphone between speaker and wall OR perpendicular to wall OR left to right OR towards the wall [could be shown by labelled arrow added to diagram] (1) Oscilloscope’trace shows sequence of maxima and minima (1) Gi) How nodes and antinodes are produced Superposition/combination interference/overlapping/crossing of emitted incident initial and reflected waves (1) Antinodes: waves (always) in phase OR reference to coincidence of two compressions rarefactions/peaks/troughs /maxima/minima, hence constructive interference ‘reinforcement (1) Nodes: waves (always) in antiphase/exaetly out of phase OR compressions coincide with rarefactions ete, hence destructive interference / cancellation (1) (8)() 28, Gi) co) fa) @) ‘Measuring the speed of sound Measure separation between (adjacent) nodes / antinodes and double to get this is ¥s/ [not between peaks and troughs] (1) Frequency known from/produced by signal generator OR ‘measured on CRO / by digital frequency meter (1) Detail on measurement of wavelength OR frequency e.g. measure several [if'a number is specified then 23] node spacings and divide by the number [not one several times] ‘OR measure several (>3) periods on CRO and divide by the number OR adjust cro so only one full wave on screen (1) Use v (allow ¢) =f. 4 Application to concert tall Little or no sound /amplitude OR you may be sat at a node (1) Sensible reason Examples: Reflected wave not as strong as incident wave OR walls are covered to reduce reflections OR waves arrive from elsewhere [reflections/different speakers] OR such positions depend on wavelength / frequency (1) 2 m1 Meaning of statement (5.89 « 10°" J/ work function) is the energy needed to remove an electron [allow electrons] from the (magnesium) surface plate Consequent mark Minimum energy stated or indicated in some way e.g. atleast/or more] (1) 2 (Calculation of time Use of P= 14.) Use of B= Pr) [use of E = Le scores both marks) Comect answer [210 (s),2 sig fig minimum, no we.) () [Reverse argument for caleulation leading to either intensity, energy or area gets maximum 2 marks} Example caleulation: 1= (5.89 « 107 J)(0.035 W 28107 mr) 3(b) c @) (b) i) wi (ii) How wave-panicle duality explains immediate photoemission owe a) Photon anzegy is hf depends on frequency / depends on wavelength (I) One/Eaeh photon ejects onelam electron (1) The (photoelastin is ejected atonce/immedtately (1) {notjust photoemission is immediate] 4 () Condition tor eflection ‘Angle of incidence greater than evtical angle [accept i> €] (1) 1 ‘Description of path ofray Any two from: + Ray meftacied at A and C + Description of direction changes at Aand C + Total internal wflectinn at B (1)(1) 2 ‘Things wrong with the diagram Angle of refraction can't be 0 / refracted too much (1) ‘No refraction on emergence from prism (1) [Allow | mark for correct reference to partial reflection] 2 Comected diagram + emergent ray roughly parallel to the rest of the emergent rays (1) + dimetion of refraction first surface correct (1) + direction of refraction second surfice correct (1) 3 (8) Caleulation of adapters input Recall of: power =V (I) Correct answer [0.01 A] 1) Example of calculation: power=/V" I= PIV=25W/230V =Q01A 2 ()— Explainwhy VA isaunit of power Power = voltage * current so unit = volt » amp (1) 1©) 3. (ii) Calenlation 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% (ii) @ Resistance (accept explanations beyond spee, e.g. eddy currents) (1) ‘hence heat loss to surroundings (1) 2 Calculation of charge Recall of O=sr(1) Correct answer [4000 C] (1) Example of calculation: Q-It =02A%6h 0.2 A» (6 * 60 * 60) 4000 € (4320) 2 Calculation of work done Recall of. = QVOR Recall of W= Pr (1) Correct substitution (1) Comect answer [13 0005] (1) Example of calculation: w-ov 320.C x 3 V [ect] 3 000 3(12 960 3) oR W=Pt \oW 6h ‘SW * (6 * 60% 60) 3 0003 3 112) (i) Addstanding waves to diagrams ‘Mark for each correct diagram (1)(1) 2(©) @ 32. () di) Gi) Mark place with largest amplitude of oscillation antinode marked [allow clear indication near centre of wave other than an X, allow correct antinede shown on diagrams B or C] (1) 1 (ii) Nameofplace marked @ (Displacement) Antinode [allow ecf from (a) (ii)] (1) 1 Calculation of wavelength Comect answer [5.6 m] Example of calculation: =2%28m =5.6m(1) 1 (ii) Calculation of frequency Recall of y=/4 (1) f=330m57/5.6m 8.9 Hz 2 Explanation of difference in sound 4s the room has a standing wave for this frequency / wavelength / it's the fundamental frequency (allow relevant references to resonance) (1) 1 ‘Suggest another frequency with explanation Appropriate frequency [a multiple of 59 Hz] [ecf] (1) ‘Wavelength 1/2, 1/3 ete (stated or used) (1) 2 Explain change in frequencies wavelengths (of standing waves) bigger / va) hence frequencies smaller‘lower (1) 2 (a) nay Bluelight: ‘Wavelength / frequency / (photon) energy 1() © @ «i @ “i Conversion of either value of eV to Joules Use of f= E/h Correct frequency range [4.8 « 10" — 8.2 « 10" Hz or range = 34 «104 Hz] {no penalty for rounding errors] eg DeV=2« 16x 10-19=3.2 x 10-197 63 x 10 xB £=48% 104 Hz BaeV= 34% 16% 10 = 5.4 107 £=8.2% 10! He 3 Diagrams: Downward arrow from top to bottom level On larger energy gap diagram 2 ‘Less heating / less energy lost / greater efficieney ! lower voltage needed / less power lost 1 Resistance: Recall of R= aLlA Use of R= pLid Conzet answer [$0(02)] [allow 80-84 (£2) for rounding errors) Eg. R=(2* 107 * 5.0 * 107) (3.0 107 «4.0 107) =o0 3 110)33. @ (®) Angles: ‘Normal correctly added to raindrop (by eye) An angle of incidence comectly labelled between normal and incident ray and an angle of refraction correctly labelled between normal and refracted ray (b) Angleof reffaction: Use of w= sin j/ sin r Correct answer [20°] [allow 20°-21° to allow for rounding errors] ex. sin r sin 279/13 0° (© @ Critical angle: ‘The angle beyond which total internal reflection (of the light) occurs [allow TLR] / r= 90° (i) Critical angle calculation: Use of w= 1 / sin Correct answer [503°] [allow 50° — 51°] Eg. Sin Diagram: i= 35° [allow 33° 37°] Ray of light shown refracting away from normal on leaving raindrop Some internal reflection of ray also shown with [by eve] Reflected ray shown refracting away ftom the normal as it leaves the front of the raindrop / angle of refraction correctly calculated at back surface© 38. © @ ) @ @ (a) ) Refiactive index: (Red light has) lower refractive index (than violet light) ‘vis (number of) charge carriers per unit volume o ‘number density or (number of charge carriers mor charge carrier density(1) [allow electrons] vis drift speed or average velocity or drift velocity (ofthe charge carriers) (1) Liust speed oF velocity seores zero) /Aand QA sor/ Cs" and OC) nm (1) Am? and vms" (1) {tf no equation written assume order is tha of equation) n Land Q Need all three Ratio va/ vy less than I following sensible calculation (1) Rat ¥4 10.25 1 1:4 (1) (ratio 4:1 scores 1) [d4vgelyy scores 1] Use of P= (1) Current in lamp A~2.A (1) [0.5 A scores zero unless 24 = J « 12 seen for 1 mark] Example of answer I=P+V=24W+12V A (i) Voltmeter reading = 12 V (1) (ii) p.d. across Ra = 6 V or their (b)(i) minus 6V (1) Use of R = VI (1) conditional on first mark R, Answer to this part must be consistent with, voltmeter reading and if voltmeter reading is wrong. this part has a max 2. If (b)(i) = 15 V then need to see If ()(i) =6V or less they are going to score zero for this section, tay 2 3 I 2 (3) 2 1 3(iii) current through Ry = 5 A (1) eef answers from (a) Example of answer Current through R; =2 A+3A=5A (iv) p.d. across Ry = 3 V (1) ecf (15V minus their (b)(i)) Example of answer p.d. across Ry = 15 V-12 V=3V. om Ri Example of answer Ri =3V+5A=06 [accept fiaction 3/5] 36. @ (Era) (i) FRA) (ii) Fry (b) EI=PR+ProrE=IR+h cect Must use values (a)()-il) (©) Mor ciceuit given by Igac= E (1 or substitution of SQQOV into the equation (1) (ior safety) need I ro be as small as possible (1)) (©) (i) Howe know the speed is constant Crest spacing constant / circular crests Or wavelength constant / equal wavelength (1) [Accept wavefront for crests] [Don’t accept wave] (ii) Caloatation of speed A is 10mm (1) [Allow 9t0 11] Useofr=f2 () 0.40. ms (1) [Allow 0.36 to 0.44 Allow last wo marks for correct ealeulation from wrong wavelength) (40r2)(10 x 10 m) 40ms* Line X 1* constructive interference line below PQ, labelled X (1) [Accept straight line Ignore other lines provided correct one is clearly labelled X] (i) Superposition along PQ Constructive interference / reinforcement / waves of larger amplitude / larger crests and troughs (1) Crests from S; and S» coincide / waves are in phase / zero phase difference / zero path difference (1) Amplitude is the sum of the individual amplitudes (OR twice the amplitude of the separate waves) (1) (i) Table38. 30. @ O) aa) Acconstructive (1) B destructive 1) 2 10) Light / Visible /red (1) 1 Calculation of work funetion ‘Use of @ = hcl (1) 3.06 » 10° [2 sig fig minimum] (1) 2 (6.63 « 1074 J3)(3.00 10° ms"")(6.5 * 10"” m) =3.06~ 10°F oO Minimum potential difference berveen C and A /aeross the photocell (1) Which reduces current to zero OR stops electrons reaching A / crossing the gap / crossing photocell (1) 2 (ii) Whythe gmphs am parallel Comet rearrangement giving ¥,= hfe ~ 9 /e (1) Gradient is ve which is constant same for each metal (1) {Second mark can be awarded without the fist ifno rearrangement is given, or if rearranged formula is wrong but ddoes represent a linear graph with gradient hie] 2 m1 (Calculate maximum cumrent Recall of P= F(A) Correct answer [0.49 A] (1) Example of calculation: Poiv I=59W/120V 4A 2(b) w (ii) Show that wesistance isabout 240 Recall of Y= JR (1) Correct answer to 3 sf [24.5 QJ [no ue] (1) Example of calculation: 2V/049A 24.50 (Calculate current Use of comrect eircuit resistance (1) Correct answer [0.45 A] (1) Example of calculation: IR 12V=(2450+20) 045A Calculate power Recall of P = IV and V'= IR (accept P= FR) (1) ops R Comeet answer [5.0 W] (1) Example of calculation: P=IR = (045 Ay x2450 =50W(© Inewasein power available io pump eg, lower resistance in wire thicker wire, panel nearer to motor (1) (accept relevant answers relating to panels, e.g. more panels) 1 1 40. (a) (i) Name process Reftaction (1) 1 Gi) Explanation of refiaction taking place change in speed / density / wavelength (1) 1 © @ Dewy fom huey fish refraction shown (1) refraction correct (1) 2 « . . a Identify the angle as that in the denser medium (1) Indicate that this is max angle for refraction OR total internal reflection occurs beyond this (1) [angles may be described in terms of relevant media] 2 (iii) Explain two paths for rays ftom fish A to fish B direct path because no change of medium/refractive index/density (1) (total internal refletion along other path / angle of incidence > eritcal angle (1) direct muy correctly drawa with arrow (1) total intemal reflection path correctly drawn with atrow (I) [lack of ruler not penalised directly] [arrow penalised ence only] 4 [10] 41. (2) Liltrasound: High frequency sound / sound above human hearing range / sound above 20 kHz ’ sound too high for humans to hear (1) I«© (@ 2. to prevent interference between transmitted and reflected signals / allow time for reflection before next pulse transmitted /t0 allow for wave to travel to be determined (1) (ii) High pulse rate: Greater accuracy in detection of prey"s motion / position / continuous monitoring / more frequent monitoring (1) 2 Size of object: Use of 4= vif (I) Correct answer (0.0049 m or 4.9 mm) (1) [accept 0.0048 m or 0.005 m] example: i= 340 m s'/ 70000 Hz 0049 m = 4.9 mm (accept 5 mm) 2 Time interval: Use of time = distance / speed (1) Correct answer (2.9 10°? s) [allow 3 « 107 s] [allow 1 mark if answer is half the correct value ie. Distance = 0.5m used] (1) example: time=1m/340ms7 2.9% 107s 2 (©) Eflieton fiequeney: Frequency decreases (1) Greater effect the faster the moth moves the faster the ‘moth moves the smaller the feequency (1) 2 1 (®) Difitastion diagsam: ‘Waves spread out when passing through a gap / past an obstale(1) stays constant (1) 2 © Diagram showing 2 waves in phase (1) Adding to give larger amplitude (1) 2© @ “Atomic spacing (similar to /) Regular / ordered structure ‘Symmetrical structure DNA is a double helix structure (2) Electron behaviour: (Behave) as waves (1) i and r comeetly labelled on diagram (1) 5 +2" (d) 34-21) [allow 1 mark if angles measured comectly from interface fe. = 65+ 2%,r=52¥- 2°] () i) Refetive index: Use of guts sin i/ sin r [allow ecf] (1) Use of gy= Vetta (I) example: gla = Sin 25 / sin 38 = 0. ‘allg= V gllg= 146 (b) Ray diagram: © @ o co) Ray added to diagram showing light reflecting at interface with, angles equal (by eye) () Obsercation: Incident angle > critical angle (1) TAR occurs (1) dargestangle sin C= 1/146 (allow eet) (1) C= sin (1/1.46)= 43° (1) ict Potential difference (1) Potential difference (1) (iii) Rateof change current (1) Max 2 1 3 2 2 10)i) (iv) Base-guantity ‘current (1) (for aay part if two answers are given score is 2ere) 4 @ ©) 14 (Astempennuce of themnistr increase) its resistance decreases [Do not credit the converse (I) any TWO (light) decrease in v (symbol, velocity or drift velocity) Large increase in n increases (accep electrons charge carriers for n] ‘A, Q and (pd) remain constant (1)(1) [ignore any reference tv staying constant) 3 (constant, can't score mark for 34) ammeter eading decreases () voltmeter reading unaltered (1) ammeter is used to indicate temperature (1) (ii) Assumption: ammeter; ideal/has zero/negligible resistance (1) (Reference to meters is zero mark) 446. 47. (a) @ Tungsten filament Qowe (1) Tis not (direetly) proportional to Temperature of filament increases! filament heats up! ‘gets hotter as current/pd increases [accept bulb or lamp but not wire] Links temperature increase to resistance increases tungsten filament does not obey Ohm’s law /not an Ohmic conductor or resistor. ()Q)) ‘Any THREE, 4 (Reading current from graph 1.5 A (1) answer 53 0 () (misread current — 0/2) Example of answer P=IR R=8041 30 2 (ii) Addition of two cuments (1) OR use of R = V/I and resistors in parallel formula 13+12=27A0) ef candidates” current from above [Ifyou see 2.7 A give 2marks] 2 (8) () Use of P= 1°/R OR P= JVand P= IR) Total R= 4.5 0(1) 2 Example of answer R=PsP=12Vx2VA2W R=450%) © (a) o) Use of UR= UR, + VR: - ORER=/SR (I) {OR find total current, divide that by 5 and use 7 =/R] Resistance of strip = 22.5 2 (1) ecf candidates” R. 2 [common error is to divide by 5+ 0.9 Q scores 0/2 but ecf to next part gives /= 0,033 m which will then score 3/3] R=pl/A or correct rearrangement (1) Correct substitution (1) Length = 0.82 m (1) ecf candidates’ R 3 Example of answer T= RAip= (259% 40% 10% m+ 11 < 10° Om 1=082m See P= I" /ROR P= IV leading to increase in current or decrease in resistance (1) ‘more strips in parallel / material of lower resistivity (D) [not greater conductivity] 2 (9) EMF. = work done / charge OR energy transferred / charge (1) OR power current [There is only one mark here and this is consistent with specification but it must not be Joules or coulombs) 1 (i) Useof Y= IRA) 20a) 2 Example of answer V/R=80V/400 20A (i) Usesp.d =4.0V @) 7 =2.0 Qeef their 1(1) 2 Example of answer V/1=40V/204 202(iii) Use of P= VI PR! PIR (A) P= 16 W ecf their /(1) 2 Example of answer P=VI=8Vx2A (b) (wv) 16 W Uses4V or2A « 20 or their <7 (1) see 5 x 60 s in an energy equation (1) energy = 24001 (1) 3 Example of answer n= 4V X2Ax5% 608 10) [Marks may be earned on diagram or in text] ‘Named light source plus polaroid (OR polariser OR polarising filter) / Laser / Named light source and suitable reflector (e.g. bench) (1) 2" Polaroid plus means to detect the transmitted light (1) (ive, eye OR screen OR LDR OR light detector OR instruction to e.g, Look through polaroids) Rotate one Polaroid [Only award if expt would work] (I) Detected intensity varies / No light when polaroids are at 90° (1) ‘Maxima and minima 90° apart / changes from dark to light every 90° (1) [Use of microwaves, slits or “blockers”: 0/5 Use of filters or diffraction gratings: lose first two marks Use of “sunglasses” to observe: lose mark 2) 5 Why sound can't be polarised ‘They are longitudinal / They are not transverse / Only transverse waves can be polarised / Longitudinal waves cannot be polarised / ‘Because the (*) is parallel to the (**) (1) o = vibration OR displacement OR escillation OR motion of particles (**) = direction of travel OR direction of propagation OR motion of the wave OR direction of energy transfer 1 (6)sh. (@) @ Table om f 24 (110) 1.2 220 08 330 All wavelengths comrect (2) (b) (a) (b) [One or two wavelengths correct gets 1] Both frequencies correct (1) [Accept extra zero following wavelength figure, e.g. 2.40. Accept units written into table, e.g. “2.4 m", “220 Hz"] 3 (i) Whynodes ‘String cannot move / no displacement / zero amplitude / ‘no oscillation / phase change of x on reflection / two waves ‘cancel out / two waves are exactly out of phase (1) (OR have phase difference of x OR halfa cycle) / destructive interference 1 Why waves with more nodes represent higher energies More nodes means shorter wavelength (1) ‘Momentum will be larger (1) [OR Allow 1 mark for*More nodes means higher frequency and 2= If") 2 (6) Why statement comect Blue photon has more energy than red photon (1) Why statement incorrect Blue beam carries less energy per unit area per second / Blue beam carries less energy per second / Blue beam carries less energy per unit area / Blue beam has lower intensity and intensity = energy per unit area per second ‘Additional explanation [Under “correet”] Blue has a higher frequency (OR shorter wavelength) / [Under “incorreet”] Blue beam has fewer photons (1) [Allow reverse statements about Red throughoat part si 3 (Meaning of work function Energy to remove an electron from the surface (OR metal OR substance) (I) [Don’t accept “from the atom”. Don’t accept “electrons”,] Minimum energy... Least energy... / Energy to just... ‘without giving the electron any kinetic energy (1)(i) Caloulation of threshold frequency Use of @ = hf (1) ‘Correct answer [6.00 x 10"4 Hz] (1) eg. (3.98 « 1071? Fy1(6.63 « 107 J s) = 6. 2 .(@)- Which transition Use of (A)E = he’ OR (AYE = lfand f= 12 (1) Use of 1.6 « 107 (1) Correct answer [1.9 eV] (1) CwoB/-150-340) Accept reverse calculations to find wavelengths] es. (6.63 « 10 15}3.00« 10° m sy" (656 = 10°? my(1.6 «10 Tev"y =19eV 4 (0) Explanation of absomption line gowe ay Light of this wavelength is absorbed by hydrogen (1) In the outer part of the Sun (OR Sun’s atmosphere) (1) Absorbed radiation is reemitted in all directions (1) Transition from B to C (OR-3.410 -1.5) (D) Max 4(o) Why galasy receding Wavelength increased (OR stretched) /red shift / frequency decreased 58. (a) Describe propagstinn of longitndinal waves Pantiles oscillate / compressions/tarefactions produced (1) oscillation’ vibration displacement parallel to direction of propagation (1) (b) Calculation of wave speed Recall of »=/2 () Correet answer [7.2 km s] (1) Example of calculation: v=fh y=9Hz x 0.8km 2 km $7 [7200 ms]©) 54. (i) Recallof v= s/r(1) Correct answer for rin minutes or hours [about 6 minutes] or relevant comment with 347 s or calculation of tidal wave speed [0.35 kms") with comment [allow eef] (1) Example of calculation: s/t = 2500 km + 7.2 kms"! OR v= 2500 km = (2 « 60 60 8) 71-3475 OR y= 0.35 kms = about 6 minutes (stated) / much less than hours / 2h is 7200s OR 7.2km >> 0.35 kms (@) Meaning of superposition ‘When vibrations disturbances/waves from 2 or more sources coincide at same position (1) resultant displacement = sum of displacements due to individual waves (1) (®) @ Explanation of formation of standing wave description of combination of incident and reflected waves! ‘waves in opposite directions (1) described as superposition or interference (1) where in phase, constructive interference / antinodes OR where antiphase, destructive interference / nodes OR causes points of constructive and destructive interference OR causes nodes and antinodes (1) Calenlate wavelength Identify 2 wavelengths (1) Correct answer [2.1 * 10”? m] (1) Example of calculation: (NANANANAN) X to Yis2 4 A=4.2% 107% m+2 2.1% 107m@ (i) ii) (ii) Explsintemns amplitude ~ maximum displacement (from mean position) (can use diagram with labelled displacement axis) (1) antinode — position of maximum amplitude OR position where waves (always) in phase (1) (i) Caleulate resistance Recall of R= (1) Correct answer [8.65 0] (1) Example of calewlation: R=vt R=268V=031A = 8650 ‘Show that internal resistance is about 0.4.0. Recall of relevant formula [1"= © J OR lost volts = (¢- 1) (2) OR ©=AR+)] including emt Comect answer [0.39 0] [no ue] [allow eet ife = AR+)] () Example of calculation: threshold frequeney / flight > fo / wavelength of light is shorter than threshold wavelength /2.< hq + 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 mode / cross the gap (i) + (Atorabove P,) 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 (ID) [Don’t 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) Effcts on the stopping poteatial () No change (1) (ii) Imoreases (1) [Ignore incorrect reasons accompanying correct statements of the effect] 66. (a) Circuit diagram and explanation ammeter and voltmeter shown in series and parallel respectively (1) current measured with ammeter and voltage / p.d. with voltmeter (1) (b) Caloulation ofsesistance Recall of R= ViT(1) Correct answer [25.0 0] (1) Example of calculation:R=VI R=300V=0.124 =25.00 (©) Calculation of resistance Recall of P= 77/R (1) Correct answer [29.4 Q] (1) 2 Example of calculation: [Accept calculation of = 8 A (1), calculation of R = 29.4.0 (1)] (4) Explanation of difference in values of resistance At higher voltage value element is at a higher temperature (1) (resistance higher because) increased lattice ion vibrations impede charge flow (more) (1) 267. (a) Explain howe vapour emits light electrons excited to higher energy levels (1) 4s they fall they emit photons/electromagnetic radiation'waves/energy (1) 2 >) (Meaning of spectaltine (when the light is split up) each frequency-wavelength photon energy is seen as a separate'discrete line (of a different colour) (1) 1 Gi) Caleulation of frequency Recall ofv=/2 (0) Comect answer [f= 5.1 10" Hz] (1) 2 Example of calculation 3.0 10? mst=fx 589 10% m f=5.1 x10" Hz (©) Explanation of different colours differen colours = different feq/wavelengths photons of diffrent energies (1) ‘photon energy frequency wavelength depends on difference between energy levels (1) Uiffatoms have diffenergy levelsidiff differences in levels (1), 3 (@ Explanation of tanswerse waves ‘aration in E or Bild oscillations/vibrations displacement aright anglesperpendicular to direction of wavel propagation [not just motion or movement for both I and 3* part} (1) 1 68. (0) _Esplanttion of maxima orminimus path difference = 2 * 125 x 10° m= 250 «10° m () = half wavelength /antiphase (1) —sdestructive interference / superposition (1) (minimum intensity)(b) Meaning of coherent remains in phase / constant phase relationship (1) 69. (a) (i) Energy level diagram: + Arrow showing electron moving from lower level to 8 higher level (1) + Arrow downvvards from higher to lower level [must show smaller energy change than upward arrow] (1) Gi) Missing energy: Causes ar semperature of a named item (1) Gi) Rangeofenergies: “Minimum energy when 4 = 400 = 10 m (1) 4.) Correot answer [3.1 eV] () [allow 3.0 ~ 3.3 eV for rounding errors] [n0 u.e] eg. f =3% 108/400 x10" =75%10"He 5.0* 107 E=3.1eV (b) Detecting forgeries: Forgery would glow / old painting would not glow (1) 70. (a) (i) Critical angle calculation: Use ofsin C= 1/ua (0) Correet answer [244° — only acceptable answer) [no u.e] (1) 2.42 eg. Sin C= Vista c=244 (i) Baydiagram: ‘Small angle ray shown passing into air, away from the normal (1) Large angle ray showing T.LR. with angles equal [by eye] (1) 4 8)) ne (a) (b) ity (iy Labelling of angles: ‘An incident angle correctly labelled between normal and ray in diamond (1) Anamgle of refraction comectly labelled between normal and ray in air (1) 2 ‘Any 3 of the following + The higher the reffactive index the greater the amount of ‘rapped light + The higher the reflactive index the lower the critical angle + TLR occurs at angles greater than the critical angle + So, iferitical angle is smaller, more light is reflected (YC) Max 3 Comment on angle: Lower ertical angle so more sparkle (1) 1 (10) Graph scale: Log scale (1) 1 a (i) Choice of materi: Any 2 of the following: + (almost) all ofthe voltage is dropped across the carbon rod + gives the greatest speed variation + others need to be longer (to have same resistance as carbon) + others need to be thinner (fo have same resistance as ‘earbou) (DA) Max 2 Use of R= pL (1) Comect units used for all terms [all in em or all in m (1) Comect answer [1.9 0] (1) 3 [allow 1.8.0 for rounding errors — no we] eg, R= 14 107° = 0.4/3.0 « 10 =199 (iii) Available voltage: X-12V Y-O0V(I) 1(iv) Less voltage available for tain set as some wasted across wires (1) 0.5 Q is (relatively) large % of total resistance, so effect is high / not negligible (1) or Calculation of potential difference available now (1) [9.5 V] [allow 9.59.6 V] Significant drop from 12 V (1) (Rey! Ryd) * Vagyy= (1.9/(1.9 + 0.5) * 12= My (91 T. (a) (i) Potential difference = work (done) (unit) change OR Potential differenc ‘i C=AsorW=Js'Q) Vekem ATS 3 (©) Converts 2 minates to 120 seconds (1) Muiplication of VIA tor VA Q (1) Energy~ 1440.3 ) 3 Example of answer: Energy =6.0V*2.0A 1208 = 1440 73. (a) n= number of charge cariers per unit volume OR number of charge carriers m* OR Sharge carrier density (1) ‘y= drift speed average velocity/drift velocity (of the charge carriers) (1) (b) mis greater in conductors /» less in insulators. (1) [There must be some comparison] larger current flows in a conductor. Dependant on having referred to n (I) 2 (statement that n large in conductor and so current large max!) (0) In series), so same current and same n and @ (1) vg greater v4, (1) valtig= 840.25) 374. (a) 8 co) © (a) (b) 16. pd-36 V0) Example of answer, pd.= 024A 15 =36V Caleulation of pa across the resistor (1) [60-36-24] Recall V=& () 1} caleulated from their pd / 41) [correct answer is 0.60 A. Common ecf'is 6Vi40 gives 1.5 A] Example of answer: N=24V/400=06A Calculation of from h-0.24 (0.36 A] (1) [allow ecf of their. common value ~ 1.26 A] ‘Substitution V= 3.6 V (1) R-1090) (— Coradiem=) r= 1.95-29 E=89-9V(I) i) 1=215-217Aq) Gi) Useof V= RQ) R=2.1-2200) (Battery or call with one or more resistive component (1) Correct placement of voltmeter and ammeter(1) (@—VaryR_ e.g variable resistor, lamps in parallel (1) Record valid readings of current and pd (consequent mark)\(1) [Do not give these marks ifthe candidate varies the voltage as well] (a) Solar Power Use of P=/zr° [no component needed for this mark] (1) Use of cos 40 or sin 50 (with or 4) (1) 2.2 [2 sf minimum, No we}(1) eg. P=1.1 «10° Wm x cos 40 * x(29 « 107 my? =22W 0)7. © () Ene Use of E=Pr(d) 18 * 10° 372.0 10°F) eg, B= 22W x (2.5% 36005) =20«105 ©) Geaph ‘Straight line with positive gradient(1) Starting the straight line ona labelled postive, (1) {Curved graphs gct 02, Straight lin below axis loses mark 2 ules that itis clearly a constaction fine (b) Work fmeti From the yincercept(1) [Accept if shown on graph] OR Given by gradient =f (or * fo) [Provided that fois marked cn their graph, or they say how to getit from the graph] (OR Read fand £, off graph and substrate into, ~ hf [Carved araph can get this mark only by use off oF equation methods] Gradient Gradient equals Planck constant(1) [Curved graph can’t get this mark] 78. (3) Wavelength 9. evo) Use of B= fA) Useofe=— (1) 1.8% 10" [2 sfminimum, No ue) (1) ce (CLS keV — (69.6 keV) « (10 «1.6 «10? keV") (6.6 = 10s = 1.64» 10 He, A = 3.00% 10 ms'/1.64 « 10" Hz =1810 m ©) De rays. [Accept gamma rays] (1) Mesning of energy level ‘Specific allowed energy/energiss (of electron in an atom{) 6) 8 (4)80. Meaning of photon Quantum /packet/particle of energy/radiation/light/electromagnetic wave(1) Eomnula dor photon enerey’ E:-E\() [Allow £; + Eton = Ex] Explanation of photon wavelengths: ‘Same energy change / same energy difference / energy the same(1) Meaning of coherent Remains in phase / constant phase relationshi() Description of sound Particles/molecules/atoms oscillate’vibrate(1) (Oscillations) parallel tofin direction of wave propagation / wave travel / wave movement [Accept sound for wave](1) Rarefactions and compressions formed [Accept areas of high and low pressure}(1)Meaning of frequency ‘Number of oseillations/eyeles/waves per second / per unit time(1) 1 Calculation of wavelength Recall -# (1) Correct answer [18 m] (1) 2 Example of calculation iG) ‘Waves reflected (at the end) (1) ‘Superposition interference of waves travelling in opposite directions(1) ‘Where in phase, constructive interference’superposition OR where antiphase. destructive interference suoemosition ‘OR causes points of constructive and destructive interference superposition (Do not penalise here if node’antinode mixed up) 3 Mark node and antinode ‘Both marked correctly on diagram (I) 1 Label wavelength Wavelength shown and labelled correctly on diagram(1) 1 Explain appearance of sting Any two fiom: * light flashes twice during each oscillation / strobe frequency twice that of string [accept light or strobe] © string seen twice during a eyele © idea of persistence of vision(2) max 2Calculate speed of waves Use ofy =r (1) Comeet answer (57m s“] (1) Example of calculation: vr =V(1.96 N /6.0 * 10° ke mv") =S12ms" Distance to aia: Use of distance = speed * tima(1) Correct answer [7.2(km) /7200(m) is the onty acceptable answer. No we.) es, Distance ~ speed * time = 3x 10 x 24 x 10° 7.2 km Why pulses are used: Any two of the following: Allow time for pulse to renum before next pulse sent To prevent interference/superposition A-continnous signal cannot be used for timing (Can’t transmit / receive at the same time(@) Doppler shift: Any three of the following (Change in frequency'wavelength of the signal [allow specified change, either increase or decrease] Caused by (relative) movement between source and observer [accept movement of aireraftiobserver] ‘Size of change relates to the (relative) speed of the aircraft [Allow frequency inereasing: do not allow frequency decreasing unless finked to aircraft moving away] Quote ve= arfia) max 2 max 3 m183. Areaofwire: Use of4 =a (1) Correct answer [1.9 * 107 (m’). Allow 1.9 x 107 and 2.0 x 107 (myJ() [Nowe] x25 «107P, 1.96 = 1077 m? Table + graph; Length / Area / x 10° mi! 0.0 05 1.0 13 2.0 25 3 3.6 a) 4. First 2 points plotted correctly to within 1 mm(1) Rest of points in straight line with origin by eye(1) Resistivity calculation: Drawn through the origin, ignoring first 2 points(1) Recall p= R (L/) [in any form] (1) Large triangle drawn on graph OR accept the use of a pair of values) read from the line [2> 3» 10° mis required in both cases] {[s-axis allowed as bottom of triangle} Comrect answer (1.2 10" © m)] () [allow 1.1 —1.3 x 10° m)] [no we] eg. , 04/34 = 10°=12* 107 Om‘hy oa he loving + Higher current lower resistance for shorter lengths/at these points + Atshorterlengths/at these points wire gets hotter * Non-uniform area‘diameter * Cable /contact resistance + Sensitivity of meters + Effect on resistance of any of the above(2) ‘Comrect diagrams showing vibrations in one plane only and in alplanes (1) | > ‘Vibrations oscillations labelled on diagrams(1) Telescope adaptation: Fit polarising filter lens (nrust be lensnot lenses} (1) At90° to polarisation direction to block the moonlight rotate until ‘cuts out moonlight (1) 85, Meaning of plane polarised (Oscillations vibrations field variations () Parallel to one ditecton, in one plane fallow line with arrow at bath ends) Doppler effect Doppler (1) If source/observer have (relative) movement [reflections off vibrating/moving atoms] (1) Waves would be bunched/compressed'stretched or formula quoted {accept diagram] (1) ‘Thus frequency / wavelength changes [accept red /biue shift(1) a) 4‘Erequency about 3 x 10" Hz Evidence of use of I-wavelength = wavenumber) laser wavenumber = 9400 or wavelength change =7.69*10" (1) “New wavenumber = 10700 [or $100] or conversion of wavelength change to m [7.69 « 10°] (1) ‘New wavelength= 935 nm (or 1240 nm] ‘Use of frequency =¢ / wavelength [in any calculation] (1) f= 3.2 » 10! Hz [note answer of 2.8» 1*= 3,34 10* = 440) Modelof light Particleiphoton/quantum model (1) Photon energy must have changed / quote E = hf(1) Energy of atoms must have changed [eredit vibrating less’more/faster/slowerI) 86 (a) (i) Lampbrightness: Lamp A (1) Langer cument through it (at 9.0 V)/greater power(1) (at 9.0 V)/smaller resistance (at 9.0 V) (i) Battery current Addition of eurrents (1) Current = 1.88- 1.92 AQ) (i) Totalsesistance R=9 VILA or use of parallel formula(l) R=46-499 () [full ecf for their current] 2 114)(>) Lampsin series 87, Current same in both lamps/current in A reduced from original value(1) Pd actoss A less than pd across BL) Lamp A has a lower resistance than lamp B(1) P=WorP=RF () Any2 Lamp A will be dimmer than B [conditional on scoring ONE of{1) 1 the above marks] 9 (a) @) Resistance Use of 7 [ignore 10°] (1) 3800.0 (3784.2) (1) 2 (i) Resistance of thermistor OR | OVE7AmA 62V=0.74mA* Rin R oR) $400.0 [8378.2] [substinuting 40000 gives 88579 ie $9002 (1) [method 2 substituting 38000 gives 8362. : substituting 4000.0 gives$162.0 } 2 (0) Suggestion and Explanation ‘The milliammeter reading increases(1) ‘Thermistor resistance “becomes zero” /Short circuit(1) ‘Since supply voltage is constant /T= 9.0 7R (1)() © (a) Definition of EME Energy (conversion) or work done (1) Per unit charge (1) OR E-mod) Symbols defined (1) [E= LC scores 1] oR £=PI() Symbols defined (1) [terminal pd when no current drawn of open circuit scores max 1) ‘Volmeter calculation Any attempt to find any current (1) Attempt to calculate pe across 102 resistor (1) S77V on Potential divider method: ratio of resistors with 10.49 on the bottom (1) Multiplied by 6.0 V() sm7Va) FFor either method, an answer of 0.23 V scores max 1] ‘Second butery added ‘Voltmeter reading increased (1) Any two of EMF unchanged Total resistance reduced current increases or “lost volts” decreases(2) [8]90. (a) oo) @ w Erequency 1.0(3) « 10"? Hz (ly Electromagnetic Spectrum (i) TR, microwave & radio in corect order above visible (1) UV with either X rays / Gamma rays / both in correct order below visible (D) (ii) Wavelength at boundary 1 «10 m/ 110m (1) Plane polarised (3) Vibrations ‘oscillations (of electric fieldvector) (1) In one direction/plane (of oscillation)(1) Description i) Digoram showing generstor labelled transmiter'generstr'source/emiter (1) ‘And suitable detector eg shows haw signal is observed by using() (mictojammeter ceo loudspeaker computer with interface [Ignore anything drawn between generator and detector but for each ark do not give credit iff grille ete is attached] “To detect max and min(1) Rotate through) 9° between mas and min (1) 4 Explanation gowca) UVited photon (1) 2 evr Ex | fv Exy>® | fu.> fimo electron can break fiee)(1) One photon absorbed by one electron 1) Both metal plate and electron are negative or repel (each other() max 2 119]on. (©) (Intensity red light inereased nothing / no discharge (1) (i) Intensity of TV inereased (Coulombmeter) discharges quicker) (© MusKE Use of E = hho (1) conversion of eV to J or vice versa ie. appropriate use of 1.6 10 (I) Subtraction fe’ ~ & [must use same nits] oF use of ill equation( 1) max KE =2.2« 10") [Candidates may convert photon energy to eV leading to max KE = 1.4 eV] Temperature increase leads to increased lattice vibrations(1) scattering flowing electrons / increased collisions of electrons (1) R= Vi {stated or implied] Q) =14V 0.198 =392 (1) ‘Discuss whetherresults support hypothesis No. Resistance is not increasing with temperature(1) . - P RQ) V-PR = 1200 W «41.9 [Mark for rearrangement OR substitution] (1) [Accept 39-41 9 J [ecf] Vv=220 Va) FR), 4 A. and use in 1200 W [Allow calculate 1 AA*V (= 220V W) [19] 18192. 93, E £4 " Electrons/atoms gain energy (1) and electrons move to higher (energy) levels(1) [Credit may be gained for diagrams in this and the next 3 parts] Explanation of how radiation emitted by meremry atoms Elections (lose energy as they) drop to lower levels(1) Emit photons / electromagnetic radiation(1) Explanation of why only certain wavelengths are emitted Wavelength (of photon) depends one energy(1) Photon energy depends on difference in enerey levels(1) Levels discrete / only certain differences / photon energies possible (1) (and therefore certain wavelengths) Different energy levels / different differences in energy level1) Calculation of charge O-ik(y =0.15 A + 20% 605 =180€@) Why micrawaves are reflected ‘Wave is reflected when passing ftom one medium to another / when density changes / when speed changes(1) Varying amplitude Any two of the following: ‘Varying differences in density ofthe two mediums produce different intensities of signal (1) Different distances travelled give different amplitudes(1) Following a reflection there is less energy available(1) 2 2 3 1 2 {10} 1 Max 2Varying time Different thicknesses of medium(1) ‘What is meant by Doppler shift Change in frequency wavelength (1) Caused by movement of a source(I) Changes due to Doppler shift ‘Wavelength increases (1) Frequency decreases (1) [Allow e.c-f. from incorrect wavelength] Any one of the following: Each wave has further to travel than the one before to reach the heart ‘¢ The waves are reflected from the heart at a slower rate(1) 94, Resistance calculation Use of R = pL/A (I) ‘Substitution R = 1.6 x 107 « 0.025 * (10) = 0.02 x (10) (1) =322 (1) Total resistance Either Section 2 = R; (16 ©) OR Section 3 = 5 * Ry (10.7 2) (1)Use of Rost “Ri + Ry + Rs (W) Regi = 58.7 [550 if30.0 used as starting point] (1) 3 lcetiifsection 3 calculated as 4 *Ry = 560 ORS25O if302 used as starting point} Why thermochromic ink becomes warm Curent produces heat reference toFR on Thermal conduction from conductive ink(1) 1 [Mark for identifying that the heating effect originates in the conduetive ink] Why only thin section transparent ‘Thinner / section 1 has more resistance(1) So even a small current will heat it Power (heating effect) given by FR / current will heat it more(1) 2 [OR opposite argument explaining why thicker section is harder to heat] 5 ; Any one of: © diffraction © reftaction io} © interference * polarisation (1) 1 What is meant by monochromatic Single colour / wavelength / frequeney(1) 1 Completion of graph Points plotted correctly [-1 for each incorrect point](1) (1) Line of best fit added across graph grid(1) 3 ‘Whatel/ tells us Maximum (1) Kinetic energy of the electrons / sm" of electrons (1) 2Thueshold stequency for sodium Correct reading from graph: 4.3 = 10" Hz (1) [Accept 4.1 = 10! = 4.7 « 10" Hz] ‘Work function Fa Wfp 66313543 «1d Hea) = 2.9 10s [Allow ecf] (1) Why threshold frequency is needed ‘+ Electron requires certain amount of energy to escape trom surface(1) ‘This energy comes ftom one photon of light 1) + B=nfay 96, Adding angles to diagram Critical angle C correctly labelled (1) Use of = 1sin C () Sin C= 1/1.09 C= 66.6" (1) Why black mark not always seen At (incident) angles greater than the critical angle(1) tui, takes place (so black mark not visible)(1), light does not reach X / X only seen at angles less thant (1) [OR opposite argument for why itis seen at angles less thanC] . . Lower jr means greater density (I) Greater density means more sugar(1)97. 98, 99, (o) Cissuits Base unit ampere OR amperes OR amp OR amps (I) Derivedquantity: charge OR resistance (1) Derived unit: ‘volt OR volts OR ohm OR ohams(1) Base quantity: current () Uf two answers are given o any of the above, both must be correct to gain the mark] (@) Toand Jupiter: Time taken forslectrors toreach Jupiter =(4.2 «10 myo 10 ms)= 14.48 s Correct substitution in v= (ignore powers often) () 1 Answer: 1448 5, 14.5 s [no ue] (I) o n= Ite 3.0 « 10° A) (1sy(1.6« 10 ©) Use of ne = Jr(1) (18-20) 10° a) © Cunent direction From Jupiter (to Io) / 0 To to the moon) (a) pd.across 4 O resistor LS (A) *4(Q) -6v() Resistance Ry Current through R: = 0.5 A (1) 6) Rosa) R-120q) {allow ecf their pd across 4.) iO} 81(©) Resistance Ry pad. across Ry = 12- 6-4 =2vQ) (Current through R, = 2. (1) -2M _ Ri FQ eo [allow ecf of pd from (a) if less than 12 V] Alternative method Parallel combination = 30.0) Circuit resistance = 12(V)2 (A) = 62.0) R-6-G+2)-190) [allow eet of pd trom (a) and R from (b)] 100. (a) Cument in filamentlamp P = Vox comect rearrangement (1) 24M ©) Gi) Sketch gph Correct shape for their axes(1) I-V quadrant showing fair rotational symmetry (1)101. @) ai) (i) Explanationaf shape (As the voltage/pd. increases), current also increases(1) (As the curent increases), temperature of amp increases) (This leads to) an increase in sesisianc of lamp(1) so equal increases in lead to smaller increases inJ OR rate of increase in curtent decreases OR correct reference to their correci(1) gradient (lf. straight line graph was drawn though the origin then(4)(0)(0)(1) for the following! Vis proportional to R ‘therefore the graph has a constant gradient) (Replacement va Explanation [ONE pair of marks] Resistance: resistance of Y; [not just the voltmeter] is muuch larger ‘than 100 © OR combined resistance of parallel combination is(1) approximately 100 ‘Voltage: p.d. across Vj is much greater than p.d. across 100 2 OR(1) all 9 V is across Vi OR Current: no current is flowing in the circuit / very small eurrent(1) Resistance: because V, has infinite/very large resistance (1) OR (Correct current calculation 0.9 x 10°* A and) correct pd calculation 90x 10* AQ) This is a very small/negligible pd(1) (®) Circuit diagram @ (wa) or equivalent resistor symbol labelled 10 MQ(1) () or equivalent resistor symbol labelled 10 MQ(1) [They must be shown in a correct arrangement with R] 4Gi) Vaeot 6 (V}: 3 (V)= 10 (MA): 5 (MA) ‘Rows of parallel combination is 51) MQ 1/3 (MO)= 1/10 (MQ) =1R OR some equivalent correct(I) substitution to show working R-10MOQ) 102, Table ‘Wavelength of light in range 390nm—700nm_| (1) Wavelength ofgamma | <1 m a Source (unstable) nuclei o ‘Type of radiation radio (waves) w ‘Type of radiation infra red w ‘Source ‘Warm objects /hot objects / | (1) above 0K 103. (a) Amplitude (by ‘Maximum distance/displacement From the mean position mid point / zero displacement line (1) cguilibrium point [If shown on a diagram, at least one full wavelength must be shown, the displacement must be labelled “a or “amplitude” and the zero displacement line must be labelled with one of the terms above.] Progressive wave Displacement at A: 2.0 (cm) [accept 2](1) Displacement at B: 2.5 (cm) to 2.7 (cmj(1) Displacement at C: 1.5 to 1.7 (em)(1) Dine [Minium] one complete simasoidal wavelength drawn(1) Peak between A and B [accept on B but not on Aj) (cm) at x = +2.6 cm with EITHER x = +6.2 cm ORx 10a) 8) 18) m104, (a) Transverse wave ine along which) particles/em field vectors oscillate/vibrate(l) Perpendicular to (1) Direction of travel or of propagation or of energy flow or velocity(1) 3 &) Standing waves Progressive waves 1. store energy 1. transfer energy (1) 2.only AN pointshavemas all have the max amplidispl (1) ampl/displ 3. constant (relative) phase 3. variable (wlative)phase relationship relationship () ‘Max 2 © © Dropless ‘Fommed at nodes /no net displacement at these points(l) 1 (i) Speed Use of = A.) Evidence that wavelength is twice node-node distance(1) ‘Wavelength = 12 ¢em)(1) Frequency = 8.0 [8.2 / 8.16] Hz or * only (1) 4 (19) 105. Photoelectic effect (3) Explanation: Particle theory: one photon (interacts with) one electron() ‘Wave theory allows energy to “build up’, ie. time delay(1) (b) Explanation: Particle theory: ft0o low then not enough energy (is released by photon to knock out an electron) (1) Wave theory: Any frequency beam will produce enough energy (to release an electron, ive, should emit whatever the frequency )(1) 2 (4106. Description of photon Packetquanturn/particle of energy [accept = lif for energy] (1) (1) {allow {packet quantum/particle} of {lighv’e-m radiation/e-m wave} ete fox) X] [zero marks if error of physics sch as particle of light with negative charge] (ay =16 «10° Cx O48V = 7.7107" 5 {no we] Calculate eificiency of phoina energy conversion Efficiency = (7.7 « 107° J= 4.0 10 J) fee} (1) 0.19 of 19% (1) 107, Explanation of pressnne nodes or antinodes Pressure constant (I) Node as a result(1) = 590 Hz) , ; r=1fd) T= 1 +590 Hz [ect] = 0.0017 sq) ‘State another frequency and explain choice e.g, $90 Hz x 2 = 1180 He (or other raultiple)(1) multiple off or comect reference to changed wavelength(1) diagram or description, e.g. NAN AN, of new pattern [ecf for A & NIU) 8 11]108. Explsin zesning of meter ‘No resistance when leads touched together'short circuit/calibration for zero erzor (1) ‘Show that resistance is about 70.0 R=VeT0) 0.84 V+ 0.0081 A = 672 {no ue]) Explain section ftompassage Other curents/voltages{resistances present(1) change in current changes reading for resistance(1) ‘Explain changes in meter reading with temperature increase Increased lattice vibrations‘ vibration of atoms'molecules(1) scattering flowing clectrons/imore collisions(1) increased resistance/inerease meter reading(1) 109. Name process of deviation Refraction (1) Completion of ray diagram B— no deviation of ray (1) ‘A and C ~ refraction of ray away from normal on entering hot air region(1) ‘A.and C ~ refraction of ray towards normal on leaving hot air region(1) Show positions of nee sunk B the same sa } [consistent with ray diagram] AandCclosertoB } (1) ‘Explanation of wobbly appearance Hot air layers rise/density varies/layers uneven(1) (Change in the amount of refraction [accept refractive index/change in direction light comes from (1) (8 (8)110. Circuit diagram _Ammeter and power souroe in series(1) ‘Voltmeter in parallel with electrodes(1) [Allow both marks if diagram shows an ohmmeter without a power pack -1 if power pack] Calulation of sistance Use of area= 9” (1) R=2.7% 107 Om <5.0 > 10' m4) =IRQ TIS Plotting graph Axis drawn with R on y-axis and labelled with units(1) Points plotted correctly [-1 for each incorrect](1) Sensible scale(1) Curve added passing through a minimum of 4 points(1) Diameter of hole Correct reading from graph ~ 0.23 mm [Allow 0.22 ~0.26 mm|(1) 111. Uinpolarised and plane polarised light Minimum of 2, double-headed arrows indicating more than 1 plane and | double-headed artow indicating 1 plane labelled unpolarised and polarised (1) ‘Vibrations oscillations labelled(1) Appearance of screen ‘Screen would look white/bright/no dark bits/light [not dark = 0](1) Explanation As no planes of light prevented from leaving screen/all light gets throughyall polarised light gets through (1) Observations when head is tilted Screen goes between being bright/no image to image/dark bits(1) Every 90°/as the polarising film on the glasses becomes parallel’ perpendicular to the plane of polarisation of the light(1) [10]Comment on suggestion Image is clear in one eye andor the other(1) If plane of polarisation is horizantal’vertical(1) oR Image is readable in both eyes(1) As the plane of polarisation is not horizontal or vertical(1) 112. How sound from speakers can reduce intensity of sound heard by driver Any 6 from: ‘graphs of 2 waveforms, one the inverse ofthe other ‘+ graph of sum showing reduced signal + noise detected by microphone ‘+ waveform inverted (electronically) + and fed through speaker + with (approximately) same amplitude as original noise + causing cancellation destructive superposition + errormicrophone adjusts amplification 113, Temperature calewlation Sx WAq) pd, across thermistor is 4.2 VQ) Current Pescensee = 930 ect their current and pa subtraction exor(1) ‘Temperature = 32 °C-34 °C [Allow eef for accurate reading](1) ‘Supply.doubled Any two from: © Current would inerease / thermistor warmss up / temperature increases ‘+ Resistance of thermistor would decrease(1) (1) + Ratio of p.s would change No OR voltmeter reading /pd across R more than doubles(1) [This mark only awarded if one of the previous two is also given] co} 6 m114. Diagram 1s. Labelled wire and a supply(1) _Ammeter in series and voltmeter in parallel(1) oR Labelled wire with no supply) Ohmmeter actoss wire(1) Readings Current and potential difference OR resistance ( consistent with diagram\(1) Length of wire (1) Diameter of wire (1) Use ofreadings R= VITOR =RAI() Awareness that A is cross-sectional area (may be seen above and credited here\(1) Repetition of calculation OR graphical method(1) Precaution Any two from: + Readings of diameter at various places ldifferent orientations + Contact errors # Zeroing instruments 4 Wire straight when measuring length + Wire not heating up / temperature kept constant(1) (1) Conductor resistance R-plia@) Correct substitution of data) 3x 1072 Re Manufacturer's recommendation ‘Larger has a lower (A) Energy loss depends on/'R / reduces overheating in wires (1) [19] 1116. Carbattery ‘Voltmeter reading: 12.2 (V) (1) 1 facia ‘Terminal pd. = 12 V + (5.0 Ax 0.042) See 12V() See 5.0 Ax 0.049 (I) Addition of terms (1) 3 Wasted power See 0.040 +0.56 © OR 28V+0.2V OR 5x (15-12) W(1) Power=15 W() 2 Efficiency (came current) 12 V/ 15 V ORPouy/Pyx = 60 WI7S W(1) Efficiency .$/80% Efficiency .8/80% (1) 2 Explanation Any two from: + Starter motor /o start car needs (very) large current E 1 Ree © Eand R Bxe0) rn noe (1) C1) (D) 2 19} 117, Wavelength 030mq) 1 Lotter A on graph A stan antinode (1) 1 Wanespeed Useofy =A naosyms ay 2 [allow ecfA = 0.15 miev =54ms") Phase relationship Inphase (D) 1 Amplitude 2.5 mm(1) 1 (6)118. Diagram 119. One arrow straight down (from-3.84 to ~5.02) (1) ‘Two arrows down (from 3.84 to~4.53, then ~4.53 to~3.02) 1) TansitionT T from —$.02 to -1.85 upwards (1) -Kinstic energy values and explanation of what has happened to lithinm atom “ineach case osevay Atom stays in-5.02 (eV) level/nothing happens to it(I) 043 eV) Atom excited to~4.53 (eV) level (1) Full credit is given to candidates who take the ke. of the electron to be 0.92 1 after collision. Any TWO correct energies with correct statement. Value of wavelength 3.9 cm—0.5 em (using interpolated sine curve)(1) = 13.4.em [accept 13.2 to 13.6 em] (I) [12.3 to 12.5 om for distance using rods(1)* ] Valuc ofamplinude Peak to peak — 4.5 em [Accept 4.3 em to 4.7 emj(]) ‘Amplitude = % = peak to peak 25 em [Accept 2.15 em to 2.35 cm] [Allow ecf for 2 mark if (1) first part shown] Calculation of frequency f= 2s =05s He) Explanation of why waves are transverse ‘Oscillation vibration displacementidisturbance at right angle(1) {o direction of propagation travel of wave) {Oscillation notin direction of wave(1)>) m1Description of use of machine t ilustrate sound wave Sound is longitudinal/mot transverse() ‘with oscillation along the direction of propagation / compressions and rarefaction) so model not helpful (1) 3 ‘1a 120. Circuit diagram Variable voltage (1) Includes ammeter and volimeter(1) in series and parallel sespectively(1) 3 [No penalty for LED bias] ‘Description of current variation in LEDs Initially, increasing voltage stil gives zero current oR Current doesn’t flow until a specific mininmm voltage(1) Current then increases... (1) swith an increasing rate of inerease(1) Noa) not proportional to oR Rrnot constant / V/l not constant / R decreases(1) 2 Calculation of resistance of green LED at 1.9 V R~ W[Stated os implied] ) =19V4 1461078 = 13002 () Calculation of power dissipated by red LED at 17'V P = IV’ [Stated or implied] (1) = 3.89 x 10° Ax 1,7 V [do not penalise mA twice] = 66x 10° W() 2 12)121. Process at A Reffaction [Accept dispersion] (1) Ray.diagram Diagram shows refraction away from normal(1) Explanation of concition in stop emergence of se light at B Angle greater than critical angle (1) Correctly identified as angle of incidence [in water|(1) Calculation of wavelength of red light in watsr =A. [stated or implied] (1) A -22x 10'mst + 42 10" He =5.24% 107 m1) 122. Difference herween polarised and unpolarised fight Polarised: vibrations in one plane (at right angles to direction of tavelX) Unpolarised: vibrations in all planes [NOT 2 planes](1) OR Contect drawing QD) Vibrations labelled (1) Meaning of advertisement (Cight vibrations are) in one plane) Evidence that glare comprises polarised light Glare is eliminated, so must be polarised Fight(l) ‘Sunglasses tured through 90° Glare would be seen through glasses(1) since they now transmit the reflected polarised light€1) (6 18)123, Charge (Charge is the current » time (1) 1 Potential difference ‘Work done per unit charge (flowing) (1) 1 Enaugy 9vx 20M) = 1800) (4) 124. Number of electrons (64 10°C), Use of ~ Qe (I) Seeing 1.6 * 10° CQ) Answer of 4.0 10" (electrons) (1) 3 [Use of a unit is ue] [eve answer: 23] Rate of flow’ (6.4 «10° C)/3.85 = 16.8/17 [nC S"] OR 16.8/17 «10? [Cs] (6.4) / 3.8 s i.e, use of! = Ot [Ignore powers of 10] (1) Conect answer [No ec-£) [1.7 or 1.68 x 10° of 1.6 = 10°") (a) 2 Unit Amp(ere)/A(D) 1 18 128, Explanation af observation Any two from: LED on reverse bias? in LED infinite) LED wrong way round so curtent is zero /LED does not conduct (very small reverse bias curent since = JR *IK=0V@)Q)126. Explanation of dimness Ry ety large / Ray much greater than Rien (1) Current very Tow / pd across LED very srl (not zero\(t) Ccuitdiagram LED in forward bias (1) ‘Variation of pd across LED (1) Voltmeter in parallel with LED alone(t) [LED in series with voltmeter 0'3] Circuit diagram “Ammeter in series with cell and variable resistor (coect symnboD(1) ‘Voltmeter in parallel with cell OR variable resistor) Power ~ voltage * custent (1) 04S Vx06A =027 Wd) Description of power output Any three from: © Current zero; power output zero/smallTow © As current increases power output also increases * Then (afterX ) power decreases © Maxinmam current; power output zero(1) (1) (1) [Accept reverse order] ean of cell oss Va) Internal resistance “lost volts” - Attempt to use == OR E= 1+ IR (1) 058V - 0.45 0.64 = 0.217 /0.29 (ay [eef an emf greater than 0.45 V1 m127. Statement Statement is false (1) Wires in series have same current (1) Use of! = nde with and e constant (1) [The latter two marks are independent] ‘Statement 2 ‘Statement is true (1) Resistors in parallel have same p(t) Use of Power= F/R leading to 7, power L(1) OR as 1 Leading to a lower valuc of FZ 3 mark consequent ‘on second 128, Description +diagram Diagram to show: Microwave source transmnitter and detector (not microphone)(1) ‘Transmitter pointing at metal plate/second transmitter from same soured) Written work to include: Moxe detector perpendicular to plateito and frahenueen /accept miler on diagram) ‘Maxima and minima detected nodes and antinodes detected(1) [Experiments with sound or light or double slit 0/4] Observation Inphase/ constructive interference ->maximum/antinode(I) Cancel outiout of phase! Antiphase’ destructive interference—rminimum node (1) How iomeasure wavelength af microwaves Distance between adjacent maxima/antinodes =A /2 (1) Measure over a large number of antinodes or nodes(1) ta (3)129. 130. 131, 5 , Use of E= iv) Use of e=/A [ignore « 10° errors] (1) =e”) For 320mm £= 3.9 (eV)and 640 nm E = 1.9 (eV) (1) 4 Photocurrent readings Work fiction of Al > 3.9 / energies of the incident photons OR threshold frequency is greater than incident frequencies(1) For Li =23 eV /f= 5.6 * 10 Hz/% = 540m hence) @ photocurrent at 320nm but not 640 nm(1) If intensity * 5 then photocurrent * 5) 3 KEy, ~-4.00/3.88 -2.30 ~ 1,7/1.58 [ignore anything with only] (1) Tee LTLS8V) 2 i) Wavelength and wave speed calculation 2 =0.96 may seeing/=2theirk (f= 2.1 Hz) (1) 2 ‘Qualitative description (Coil) oscitlates / vibrates) With SHM / same ficquency as wave (cir value)(1) Parallel to spring /dicection of wave() 3 8 Electrons excited to higher energy levels) as they fall they emit photons /radiation(1) 2 [Accept 21 cm line arises from ground state electron changing spin orientation (2) / relative to proton (1)]Photon frequency related to energy /E =f ) Energy of photon = energy difference between levels Jf Ey ~ E> (1) Energy levels discrete/quantised / only certain energy differences possible(I) ‘Shou: that hydmgen frequency comesponds to = 21cm P= 44623 «10 + Aad Hz) 2 =3 x10" +142 «10° He) 2 =0.211 mor 211m [noup] (1) 132, Explanation of assumption that volimeter does not affect values ‘Voltmeter has very high resistance takes very small current(1) Cument through X ASA+O=08A OR 48 V+609 =0.8A (1) Value missing fiom E7 Pel P244a%53V=233 WO) Description ofanpearance of lamp X as lamps switched on Gets dimmer from table, voltage decreasing / current in X decreasing / power per lamp decreasing) So P decreases (I) ‘Eomula foreell C6 THB / Rel) 1=120/(15 + B6)Q) Eiiictof internal resistance on power Power has a maximum value(1) when external resistance = intemal resistance(1) 3 [19]133, Fundamental frequency of note 440 Hz (0) Exequencies of first three overtones 880 Hz 1320 Hz 1760 Hz ‘Two comrect frequencies (1) ‘Third correct frequency (I) ‘Comment on the pattem Any 2 ftom the following: [Allow ecf] 880 Hz=2 x 440 Hz 1320 Hz = 3 » 440 Hz 440 Hz 2» 880 Hz (1) () [OR They are multiples(1) of the fundamental (or similar qualification)(1)] [Allow 1 mark for amplitude decreasing with frequency] ‘Measurement of period Example: 7 eycles takes (0.841 0825) s_ [at least Seycles](1) Period = 0.016 s +7 =23%10%s [inrange 2.2 10s to 24x 10° s](1)1M, Calculation of frequency f=1TQ) 122% 10's [Allow ecf] 54 Hz (1) Mark ondiagam Comrectly drawn normal (1) Correctly labelled angles to candidate’s normal(1) Show shat refractive: inde of water is about 13 Angles correctly measured: ins3@2y 392°) sini sin r= sin 53° /39° = 1.27 [Allow ecf] [Should be to 2 dip. min)(1) Critical angle b= Usinc 0) so sin C= 1/1.27 80 C = 52° [eet] (1) [use of 1.3 gives 50°] Explanation of reflection of ray Internal angle of incidence ~ 39° 1°(1) Compare with critical angle (1) ‘Valid conclusion as to internal reflection being total‘partil(3) Refinctive index Ic-varies with colour() (19)138, 136. Esplanations (Refraction: eg. bending of wave when travelling from one medium to another [OR change of speed](1) (ii) Diffraction: ce, spreading of wave when it goes through a gap(1) Diagram of wavestonts near beach Gradual bend in wavefronts (1) ‘Smaller wavelengtns (1) ‘Waves bending upwards as they approach shore(1) Diagram of waveftonis in bay Constant wavelength (1) Waves curve (1) Esplanation Refraction/diffraction causes waves to bend towards the beach(1) Measurement needed Any three from: © Resistance * Distance between probes © Effective area/cross sectional area R=p 4 aaa) Equation of line A Intercept= 3.5 (0 m) (+403) (1) Gradient= 1.5 @ mm) (+005) (I) So equation isp = 15d~3.5 [Or equivalent, e.cf allowed] (1) (8)137. Addition of tine Points correctly plotted (-1 for each error, allow Line of best fit drawn (1) Best distance Benveen 1,90 and 1.99 km(1) Ultasound Utrasound is very high fequeney sound(t) ‘Howulrasound can beused Any three from: + gel between probe and body + ultrasound reflects time taken to reflect gives depth of boundary * probe moved around to give extended picture # size of reflection gives information on density different(1) (1) (1) 138. ‘How reflected ullrssound provides information about heart Any vo from: + Doppler effect ‘+ frequency changes ‘+ when reflected from a moving surface ves speed of heart wall + sives heart ate (1) (2) Requires 9 V bates Battery required for electronic circuitry microphone / speaker(1) Rubberized foam ear cups: Air filled material / material has large surface area(h) Air molecules collide frequently with material(t) 119] (8Foam deforms plastically/collisions are inelastie(1) Sound converted to heat in material (1) Active noise attenuati Noise picked up by microphone (1) Feedback signal inverted / 180° out of phase with noise / antiphase(1) Amplified [OR amplitude adjusted] and fed to earphones / speaken(1) ‘Sound generated cancels Superimposes/minimum noise(1) Diagrams of superposing waves showing (approx.) cancellation(1) Amplifier gain automatically adjusted if noise remains(1) Device only works over frequency range 20— 800 Hz(1) Max 6 ‘Where does the enerey go? Some places will have constructive interference(1) More intense noise (1) Some noise dissipated as heat in air / foam(1) increased kinetic enerev of air [OR foam! molecules(1) Max 2 139, Resistance calculations Evidence of 200 for one arm (1) ieee 4-4.5¢ R730 °30 R-109() 3 Comment This combination used instead of a single 1002 resistor [or same value as before] ) because a smaller current flows through each resistor‘reduce heating in any one resistoraverage out crrors in individual resistors(1) 2 (5)140. Graphs Diode: RH quadrant: any curve through origin 1) Graph correct relative to labelled axes(1) LH side: any horizontal line elose to axes(1) RH quadrant: Any curve through origin (1) Curve correct relative 0 axes(I) LH quadrant: Curve comrect relative to RE quadrant(1) [Ohmic conductor scores 0/3) AML, Cizcuit “Ammeters and two resistors in series(1) [1 mask cireuit penalty for line through cell or resistor] Cellemf E= 130 x 10° (A) x 40 x 10° (Q) total R (A) Powers of 10(1) E-60(¥) ‘Nenccinnit ‘Voltmeter in parallel with25 (K2 ) resistor (1) ‘Resistance of voltmeter 60) (Tota resistance) = 75 gay iG}= (35.3 kQ) (Resistance of Il combination) = 35 ~ 1512 = (202) [e.c. their total resistance] Ry = 1002 [108 k2 if Ry calculated correctly] Alternative route 1: ped. actoss 15 ke (-. pad across Il combination = 3.45 V) resistance combination = 20 Ry = 1000, Alternative route 2: ‘pd. actoss parallel combination = 3.45V Through 25 kQ = 138) A SR, = 100K a a a @ @ a @ @ @142, Resistance of strain gauge a saier =P) a) Use of formula (1) x6) 0.13 @ [eof their /] () pa! 99x10" Om x24 x10 m6 4 1x10 29.6x107 130 Change in resistance A R-0132 x0.001 AR=13%104(Q) [noect} oR A R= 002 x0.001 AR=20«10°2 AR=20%10°2 0.1% 40.001 (1) Correct number for A R (1) Diiftvelociny Stretching causes to increase (1) ‘Any two from: * Current will decrease nd O Drift velocity » decreases ‘nde constant (1) (I) ‘ede constant]143, Explanation Clarity of written communication(1) Wave reflects off bench (I) (Incident and reflected) waves supespose’stationary wave is formed(1) “Maxima or antinodes where waves in phase or constructive interference occurs (1) ‘Minima or nodes where waves exactly out of phase or destructive interference occurs (1) Speed of sound See a vale berween 5.0 and 5.6 (cm)(l) Useofe =A (1) 2, =2 spacing) 320m s* to 360ms* () Explanation ofcontrast ‘AS height increases, incident wave gets stronger, reflected wave weakei(1) ‘So cancellation is less effective [consequent mark) 14. Jonisaion energy dose) «enue se-hay (1.66 «10% ay Kinetic energy eae Transition Use of B= He’ (1) 39) Transition is fram (-)1.6 eV (95.5 eV 148, Deductions abou incident radiations (@) — Radiations have same frequency/same wavelengtly same photon energy (1) GG) Tvensity is greater in (a) than in ) (1) 111]Sketch graph (c) Line of similar shape, starting nearer the origin.an negative’ axis (1) Maximum speed UseotE = 17) Subtmet 7.2 « 10" (ya) Equate to mv? (1) 3.1 <10° ms") 146, Wavelength ut. Distance between two points in phase(1) Distance between successive points in phase(1) [May get both marks from suitable diagram] Sunbum more likely om UY LUV (photons) have more energy than visible light (photons) Since shorter wavelenath /higher frequency(I) What happens toatoms ‘Move up energy levels/excitation onization(4) Correcily related to electron energy Tevels(1) Emitted pulse Greater amplitude/pulse is largeritaller(1) Depth ofl 2d=ve= 5100m -024m Henced = 0.12 m ‘Reading from graph [4.8 or 48 only](1) Ta 4.g 108s 8Calculation of 2d [their reading » timebase « $ 100}(1) Halving their distance (1) Description of trace A reflected peak closer to emitted/now 3 pulses(1) Exact position eg. 1.6 em from emitted(1) Diagram Shadow region (1) Waves curving round erack (1) 148. Resistance in darkness In the dark R =4kQ (1) so resistance per mm = 40002/40 mm = 100.2 (mm) (1) Resistance of Simm length In the light = 200 Q (1) so resistance of 8 mm strip = (8 mm/40 mm) x 2000 [= 409 } (1) Calculations Resistance of remainder = 32 mm = 1009 mnt (Total resistance = 32400 (1) Current = R= 1.2.V/3240.0 =3.7 107 A) 32002 «ay Gi) Across 8 mm, pd. =/R= 3.7 «107A» 400 () =001sVQ) Explanation of why cument decreases Any 10 points from: ‘+ more of strip is now in the dark ‘© greater total resistance #1 PRwhere Fis constant(1) (I) Max 2 (3) my149, Ermae inciccnit dingcam Cell needs to be reversed(1) ‘Any one point from: ‘© electrons released from the magnesium ‘© copper wire needs to be positive to attract electrons(1) Completion of sentence UV is made up of particles called photons(1) 1 UV and visible light (UV has shorter wavelength higher frequency higher photon energy (1) Gil) Both electromagnetic radiation both transverse waves Same speed (in vacuum) (1) ‘ stencil . t Any three points from: © reference to photons or E = hf * frequency > threshold frequency * electron must have sufficient energy to be released * UV photons have more energy # electron is released by ONE photon © brighter fight just means more photons(1) (1) (1) Max 3 Why current topped Glass prevents UV reaching magnesium(1) 1 (1 180, Total intemal reflection Any two points from: ‘+ from a more dense medium to a less dense mediunv/high to low reftactve index ‘incident angle greater than the critical angle ‘+ Tights flected nor refiacted/no light emerges(1) (1) Max 2 Critical angle Sin 1 siny =p ;givessin 90°%inC= (1) cara) 2Diteenan Reflection (TLR) at top surface (air gap)(1) Reflection (TTR) at bottom surface and all angles equal by eye(1) Path ofray A Passing approximately straight through plastic into glass(1) Emerging at glass-air surface (1) Reftaction away from normal (1) Why there are bright and dark patches on image Bright where reffacted reference to a comect ray A in lower diagram(1) Dark where air gap (produces TIR)/reference to correct top diagram(1) 151. Polarisation The (wave) oscillations (1) occur only in one plane(1) [OR shown with a suitable diagram] ‘How to measure angle of rotation Any four points from: © Polaroid filter at one/both ends * with no sugar solution, crossed Polaroids (top and bottom of tube) block out light © sugar solution introduced between Polaroids * one Polaroid rotated to give new dark view + difference in angle between two positions read from scale(1) (1) (1) (1) Graph Points ploted correctly [-1 for each incorrect; minimum mak 01) (1) ‘Good best fit line to enable concentration at 38° to be found) Concentation 0572001) kel” 182, Explanations of observations ‘Speed of light is much greater than speed of sound(1) ‘Speed of sound in soil is greater than speed of sound in ai(1) (11) Max 4 (19) 21st. 12 Vx12V cow RMT R=242 Lamp A: resistance of A decreases with current increase Lamp B: resistance of B increases with current increase Dim filament Lamps are dim because p.d. across each bulb is fess than 12 V Why Glament of lamp A is brighicr Bulbs have the same current pd. across A> pad across Bitesistance A> Resistance B oR power in A> power in B S00W n= 22 nossay 500 In22A Drift velocity greater in the thinner wire / toaster filament 181Explanation Quality of writen communication SeeT= niov is the same (at ll points) (probably) m (and ) isthe same in both wires 1 A a2 easton ma Re (ie, product =0 (6.0x10Om) x (8x10 ~ (3x10 m)x (0.00110 m) OR _ (6.0x10*Qm)x(8mm) ~ (.0mm)(1.0x 107m) R=1602