gradient of a graph or in uncertainty calculations. The MS will clearly identify the number of significant figures required. 4. Calculations 4.1 Bald (i.e. no working shown) correct answers score full marks unless in a ‘show that’ question. 4.2 If a ‘show that’ question is worth 2 marks. then both marks will be available for a reverse working; if it is worth 3 marks then only 2 will be available. 4.3 use of the formula means that the candidate demonstrates substitution of physically correct values, although there may be conversion errors e.g. power of 10 error. 4.4 recall of the correct formula will be awarded when the formula is seen or implied by substitution. 4.5 The mark scheme will show a correctly worked answer for illustration only. 5. Quality of Written Communication 5.1 Indicated by QoWC in mark scheme. QWC – Work must be clear and organised in a logical manner using technical wording where appropriate. 5.2 Usually it is part of a max mark, the final mark not being awarded unless the QoWC condition has been satisfied. 6. Graphs 6.1 A mark given for axes requires both axes to be labelled with quantities and units, and drawn the correct way round. 6.2 Sometimes a separate mark will be given for units or for each axis if the units are complex. This will be indicated on the mark scheme. 6.3 A mark given for choosing a scale requires that the chosen scale allows all points to be plotted, spreads plotted points over more than half of each axis and is not an awkward scale e.g. multiples of 3, 7 etc. 6.4 Points should be plotted to within 1 mm.  Check the two points furthest from the best line. If both OK award mark.  If either is 2 mm out do not award mark.  If both are 1 mm out do not award mark.  If either is 1 mm out then check another two and award mark if both of these OK, otherwise no mark. For a line mark there must be a thin continuous line which is the best-fit line for the candidate’s results. Question Number Answer Mark 1 D power kg m2 s−3 1 Incorrect Answers: A – Coulombs is not an SI base unit B – Incorrect, as the unit for charge in SI base units is A s C – J s1 is not in SI base units 2 B 1 Incorrect Answers: A – this is the trajectory for a ball kicked at an angle greater than , at a greater speed C – this is the trajectory for a ball kicked at an angle greater than , at a greater speed D – this is the trajectory for a ball kicked at an angle lower than , at a lower speed 3 A amplitude of the vibrations of the lattice ions 1 Incorrect Answers: B – the distance travelled by charge carriers between collisions will decrease if the temperature increases C – the drift velocity of conduction electrons will decrease if the temperature increases D – the number of conduction electrons per unit volume will remain constant if the temperature increases 4 B filament bulb 1 Incorrect Answers: A – this is not the graph for a diode C – this is not the graph for an ohmic resistor D – this is not the graph for a thermistor 5 C ± 0.1 % 1 Incorrect Answers: A – the calculation has not been multiplied by 100 to give the % uncertainty i.e. ଴.ଵ ଽଷ = 0.001 B – the uncertainty in mm has not been converted to cm and the calculation has not been multiplied by 100 i.e. ଵ ଽଷ = 0.01 D – the uncertainty in mm has not been converted to cm i.e. ଵ ଽଷൈ100 = 1 6 C 190 kJ 1 Incorrect Answers: A – The velocity was not squared when using the formula Ek = ½ mv2 e.g. ½ (1.2  103)(18) = 11 kJ B – The velocity was not squared and the ½ was omitted when using the formula Ek = ½ mv2 e.g. (1.2  103)(18) = 22 kJ D – The ½ was omitted when using the formula Ek = ½ mv2 e.g. (1.2  103)(18)2 = 390 kJ 7 A 1 Incorrect Answers: B – the ammeter would measure the current in the cell, but the voltmeter would not be measuring the p.d. across the cell C – the voltmeter would measure the p.d. across the cell but the ammeter would not be measuring the current in the cell D – the voltmeter would measure the p.d. across the cell but the ammeter would not be measuring the current in the cell 8 C 42 m 1 Incorrect Answers: A – 141 m is ¾ of the internal circumference of the track (¾  2    30 = 141 m) B – 141 m is ¼ of the internal circumference of the track (¼  2    30 = 47 m) D – 30 m (the radius) is the displacement travelled in one direction (downwards from the start position) (Total for Multiple Choice Questions = 8 marks) (Total for Question 9 = 6 marks) Question Number Acceptable answers Additional guidance Mark 9(a)  A statement applying the conservation of energy to the circuit  Use of Ohm’s law for each term /individual pd leading to the cancelling of currents in the equation  RT = R1 + R2 + R3 (1) (1) (1) Example of calculation ε = V1 + V2 +V3 ε = IRT = IR1 + IR2 + IR3 RT = R1 + R2 + R3 3 9(b) Either  Use of equation(s) to determine the total resistance of the voltmeter and 40 Ω resistor in parallel (34.3 Ω) i.e. potential divider formula or Ohm’s law  Use of ଵ ோ೅ൌ ଵ ோభ൅ ଵ ோమ  Rv = 240 (Ω) Or  Use of Ohm’s law to determine the current through voltmeter (0.0075 A) i.e. current in 40 Ω resistor calculated (0.045 Ω) and subtracted from current in 80 Ω resistor (0.0525 Ω)  Use of Ohm’s law with 1.8 V to calculate the resistance of the voltmeter  Rv = 240 (Ω) (1) (1) (1) (1) (1) (1) 1 34.29 Ω ൌ 1 40 Ω ൅1 ܴ௏ Example of calculation ቀ ோ ଼଴ ஐାோቁ 6 V = 1.8 V R = 34.29 Ω RV = 240.2 Ω 3 (Total for Question 10 = 7 marks) Question Number Acceptable answers Additional guidance Mark 10(a)  Weight of the picture is equal to the vertical component of tension  mg/2 = Tcosθ where θ is the angle between the wire and the vertical  As the angle (to the vertical) is the smaller in arrangement 1, the cosine of the angle will be larger  Arrangement 1 as the tension in the wire is lower than in arrangement 2 (1) (1) (1) (1) MP1: may be implied in an equation, accept mg = Tcosθ MP1/2: mg/2 = Tcosθ with θ defined scores 2 marks MP2/3: the angle used must be defined in words or on the diagram. Answer can be in terms of sin θ if θ defined as the angle between the wire and the horizontal e.g.MP2: mg = 2Tsinθ MP3: angle larger in 1 so sine of angle is larger MP4 conditional MP2 or MP3 4 10(b)  The weight does not act through the nail/pivot Or the centre of gravity is not in line/below the nail/pivot Or there is a perpendicular distance between the weight and the nail/pivot  There is now a moment of the weight Or the anticlockwise moment is greater than the clockwise moment  The idea that the picture stops moving when the c of g is below the nail (1) (1) (1) Accept centre of mass for centre of gravity (Allow annotations to a diagram with additional explanation for MP1/3) MP3 Accept: the turning moment being 0 Or the clockwise moments equal to the anti-clockwise moments 3 (Total for Question 11 = 8 marks Question Number Acceptable answers Additional guidance Mark 11(a)  Use of v = s/t for the horizontal motion Or see uh = 25 m s−1  Use of s = ut + ½ at2 with s = 0 Or v = u + at with t = 1.0 s Or see uv = 9.81 m s−1  Combining of horizontal velocity and vertical velocity expressions Or see tanθ = ቀ ଽ.଼ଵ ଶହቁ  θ = 21° (1) (1) (1) (1) Example of calculation horizontal motion: uh = ହ଴ ୫ ଶ.଴ ୱ= 25 m s−1 vertical motion: 0 = (uv) (2.0 s) + ½(− 9.81 m s2)(2.0 s)2 uv = 9.81 m s−1 tanθ = ௨౬ ௨౞ = ଽ.଼ଵ ୫ ୱషభ ଶହ ୫ ୱషభ θ = 21.4° 4 11(b)  Construction of vector diagram with 2 N/weight and 9N / catapult force labelled and all three directions shown  Correct scaling of 9 N and 2 N forces  Magnitude = 7.6 N to 8.0 N  Direction = 27° to 31° (1) (1) (1) (1) MP2: Award if MP3 awarded. Otherwise, the ratio of the lengths should lie between 4.3 and 4.8 (if no diagram, only MP3/4 can be awarded if answers obtained by calculation) 4 9.0 N 2.0 N (Total for Question 12 = 7 marks) Question Number Acceptable answers Additional guidance Mark 12(a)  Use of p = mv  Use of principle of conservation of momentum  Magnitude of velocity = 0.2 m s−1 with direction to the left (1) (1) (1) MP1: see 0.3 m, 0.7 m or 2mv MP3: accept ‘in the initial direction of glider 2’ for ‘to the left’ Example of calculation (taking the initial direction of glider 1 as positive) 0.3 m – 0.7 m = 2mv v = − 0.2 m s−1 3 12(b)  Glider 1 exerts this force on glider 2, so according to N3  Glider 2 will exert an (equal and) opposite force on glider 1  There is now a resultant force on glider 1  Glider 1 accelerates according to N1 Or glider 1 now moves to the left according to N1 (1) (1) (1) (1) 4 (Total for Question 13 = 6 marks) Question Number Acceptable answers Additional guidance Mark 13(a)  The (voltmeter) reading will increase  Resistance increases (with length) Or resistance  length (1) (1) MP2: accept idea of a potential divider i.e. the ratio of the of BC to the total length AD will be greater, so the proportion of the total voltage will be greater ( ୆େ ୅ୈ V) MP2: Do not award if there is also a reference to resistivity increasing 2 13(b)  Use of V = IR  Use of R = ρl/A  (Min) resistivity = 160 (Ω m) Or (max) resistivity = 730 (Ω m)  Compacted clay pathways and limestone are present in the soil (1) (1) (1) (1) Example of calculation Rmin = ଵ.଼ ୚ ଽ.ହ ൈଵ଴షయ ୅ = 189.5 Ω Rmax = ଼.଴ ୚ ଽ.ହ ൈଵ଴షయ ୅ = 842.1 Ω ρmin= ோ஺ ௟ = ଵ଼ଽ.ହ  ൈ଴.଺ହ଴ ୫మ ଴.଻ହ ୫ = 164.2 Ω m ρmax= ோ஺ ௟ = ଼ସଶ.ଵ  ൈ଴.଺ହ଴ ୫మ ଴.଻ହ ୫ = 729.8 Ω m conclusion to be consistent with calculated values 4 Question Number Acceptable answers Additional guidance Mark 14(a)  Attempt to find area under the graph  Length from 18 000 m to 20 000 m  Comparison of calculated value to 23 km e.g. The length is long enough (1) (1) (1) MP1: use of triangles or counting squares MP3: conclusion to be consistent with calculated value Example of calculation Area under the graph (counting large squares) = 18.7 × 100 m s1 × 10 s = 18 700 m 3 14 (b)(i)  26 - 28 s (1) Unit required 1 14(b)(ii)  Use of gradient of the graph between 28 and 46 s  Acceleration = 16 - 17 m s−2 (1) (1) Example of calculation Gradient of tangent = ସଽ଴ ୫ ୱషభି ଴ ୫ ୱషభ ହଶୱିଶଶୱ Acceleration = 16.3 m s−2 2 14(b)(iii)  Use of ΣF = ma using a from (ii)  ΣF = (89 + 120) × 103 N − frictional force  Frictional force = 80 kN to 84 kN (full ecf for acceleration) (1) (1) (1) Example of calculation (89 + 120) × 103 N - F = 7790 kg × 16.3 m s−2 F = 82.0 × 103 N 3 (Total for Question 14 = 15 marks) *14(c) This question assesses a student’s ability to show a coherent and logically structured answer with linkages and fully-sustained reasoning. Marks are awarded for indicative content and for how the answer is structured and shows lines of reasoning. The following table shows how the marks should be awarded for indicative content. Number of indicative marking points seen in answer Number of marks awarded for indicative marking points 6 4 5 - 4 3 3 - 2 2 1 1 0 0 Indicative content  At greater speed, the drag force is greater  Resultant force decreases Or acceleration decreases  When the rocket is started the (resultant) force/thrust increases  The mass/weight of the car decreases as fuel is used up  Increasing the acceleration (for a given applied force)  When the brakes are applied, there is a deceleration Or when the brakes are applied the resultant force is in the opposite direction Or deceleration (due to drag forces) decreases due to smaller drag forces acting on the car at lower speeds The following table shows how the marks should be awarded for structure and lines of reasoning Number of marks awarded for structure of answer and sustained line of reasoning Answer shows a coherent and logical structure with linkages and fully sustained lines of reasoning demonstrated throughout 2 Answer is partially structured with some linkages and lines of reasoning 1 Answer has no linkages between points and is unstructured 0 6 Question Number Acceptable answers Additional guidance Mark 15(a)  No repeats Or range too small Or readings should be repeated (and mean calculated) Or separations less than 0.50 m should be used Or separations greater than 1.00 m should be used Or distances should be recorded to the nearest mm Or smaller intervals (for the separation) should be used (1) Do not credit: more readings should be taken references to the number of decimal places 1 15(b)(i)  Axes labelled with quantities and units  Suitable scale  Correct plotting Microphone separation/ m Time interval / ms 1.00 3.2 0.90 2.8 0.80 2.4 0.70 2.1 0.60 1.9 0.50 1.5  Line of best fit (1) (1) (1) (1) Graph may be landscape and axes either way round MP2: Suitable scale – plotted points take up at least half the graph paper in both directions 4 0.0 0.2 0.4 0.6 0.8 1.0 1.2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 Microphone separation / m Time interval / ms (Total for Question 15 = 10 marks) 15(b)(ii)  Use of gradient Or use of 1/gradient (if axes swapped)  Speed of sound = 280 - 310 m s−1 (1) (1) MP1: only award for a y/x calculation if the graph passes through the origin Example of calculation Speed of sound = ଴.ଽ଺୫ି଴.଴ସ୫ ሺଷ.଴ି଴.଴ሻൈଵ଴షయ୫ = 307 m s−1 2 15(c)  Value obtained is lower than accepted value Max 2  Measured times are too small  Large percentage uncertainty for time  Parallax when reading from the metre rule  The distance not measured to same position on each microphone (1) (1) (1) (1) (1) MP1: accept answer consistent with calculated value but do not accept different for greater/lower Ignore references to temperature 3 Question Number Acceptable answers Additional guidance Mark 16(a)(i)  Oscillations/vibrations (of the light) are in one plane only  Plane includes the direction of energy transfer Or Plane includes the direction of travel/propagation (1) (1) Accept:  Oscillations/vibrations (of the light) are in one direction only  perpendicular to the direction of propagation/travel Or perpendicular to the direction of energy transfer Allow labelled diagrams for each marking point 2 16(a)(ii)  The (angle of polarisation of the) filters are 90° to one another Either  If plane of polarisation of light is rotated (by 90°) when it passes through the crystal (with no p.d. across it), it can still pass through the upper filter Or  If plane of polarisation of light is not rotated (by 90°) when it passes through the crystal (with a p.d. across it), it cannot pass through the upper filter (1) (1) (1) MP2: it must be clear as to whether the candidate is describing a light screen or a dark screen 2 16(b)(i)  Use of I = P/A  P = 0.014 W (1) (1) Example of calculation P = 7.8 W m−2 × 1.8 × 10−3 m2 P = 0.014 W 2 16(b)(ii)  Use of P = VI  Use of efficiency = ୳ୱୣ୤୳୪ ୮୭୵ୣ୰ ୭୳୲୮୳୲ ୲୭୲ୟ୪ ୮୭୵ୣ୰ ୧୬୮୳୲  Efficiency = 0.19 or 0.20 or 19 % or 20 % (1) (1) (1) Example of calculation Power input into LED = 3.6 V × 20 × 10−3 A = 0.072 W Efficiency = ଴.଴ଵସ ୛ ଴.଴଻ଶ୛ = 0.194 ecf from (b)(i) for the power output of LED 3 (Total for Question 16 = 13 marks) 16(c)  X is brittle at greater stresses/forces  Y will deform plastically at greater stresses/forces  The Young modulus for X is greater than Y  A screen made from material Y would be more suitable as it is more flexible (1) (1) (1) (1) Accept converse for MP3 and MP4 MP4: accept less stiff for flexible. 4 Pearson Education Limited. 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