2007 Physics. Higher. Finalised Marking Instructions
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1 007 Physics Higher Finalised Marking Instructions Scottish Qualifications Authority 007 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is to be used for any other purposes written permission must be obtained from the Assessment Materials Team, Dalkeith. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre's responsibility to obtain the necessary copyright clearance. SQA's Assessment Materials Team at Dalkeith may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.
2 Scottish Qualifications Authority Detailed Marking Instructions Higher Physics 1. General Marking Instructions SQA published Physics General Marking Instructions in July Please refer to this publication when interpreting the detailed Marking Instructions.. Recording of marks The following additional advice was given to markers regarding the recording of marks on candidate scripts. (a) (b) (c) (d) (e) (f) (g) The total mark awarded for each question should be recorded in the outer margin. The inner margin should be used to record the mark for each part of a question as indicated in the detailed marking instructions. The fine divisions of marks shown in the detailed Marking Instructions may be recorded within the body of the script beside the candidate s response. Where such marks are shown they must total to the mark in the inner margin. Numbers recorded on candidate scripts should always be the marks being awarded. Negative marks or marks to be subtracted should not be recorded on scripts. The number out of which a mark is scored should never be recorded as a denominator. (½ mark will always mean one half mark and never 1 out of ) Where square ruled paper is enclosed inside answer books it should be clearly indicated that this item has been considered by the marker. The mark awarded should be transferred to the script booklet inner margin and marked G. The mark awarded for each question should be transferred to the grid on the back of the script. When the marker has completed marking the candidate s response to all questions, the marks for individual questions are added to give the total script mark. The total mark awarded for an individual question may include an odd half mark ½. If there is an odd half mark in the total script mark, this is rounded up to the next whole number when transferred to the box on the front of the script. Page
3 3. Other Marking Symbols which may be used TICK Correct point as detailed in scheme, includes data entry SCORE THROUGH Any part of answer which is wrong. (For a block of wrong answer indicate zero marks.) INVERTED VEE A point omitted which has led to a loss of marks. WAVY LINE Under an answer worth marks which is wrong only because a wrong answer has been carried forward from a previous part. G Reference to a graph on separate paper. You MUST show a mark on the graph paper and the SAME mark on the script. 4. Marking Symbols which may NOT be used. WP Marks not awarded because an apparently correct answer was due to the use of wrong physics. ARITH Candidate has made an arithmetic mistake. SIG FIGS or SF Candidate has made a mistake in the number of significant figures for a final answer. Page 3
4 Physics Marking Issues The current in a resistor is 1 5 amperes when the potential difference across it is 7 5 volts. Calculate the resistance of the resistor. 1. V=IR 7 5=1 5R R=5 0Ω Answers Mark +comment Issue (1) Ideal Answer. 5 0Ω () Correct Answer GMI (1½) Unit missing GMI (a) Ω (0) No evidence/wrong Answer GMI 1 5. Ω (0) No final answer GMI 1 V R= = =4 0Ω (1½) Arithmetic error GMI 7 I 1 5 V 7. R= =4 0 Ω Formula only GMI 4 and 1 I 8. R= I V = Ω Formula only GMI 4 and 1 V R= = = Ω (1) Formula + subs/no final answer GMI 4 and 1 I 1 5 V R= = =4 0 (1) Formula + substitution GMI (a) and 7 I 1 5 V R= = =5 0Ω Formula but wrong substitution GMI 5 I 7 5 V R= = =5 0Ω Formula but wrong substitution GMI 5 I 1 5 I R= = =5 0Ω (0) Wrong formula GMI 5 V V=IR 7 5 = 1 5 R R=0 Ω (1½) Arithmetic error GMI V=IR I 1 5 Formula only GMI 0 R= = =0 Ω V 7 5 Page 4
5 007 Physics Higher Marking scheme Section A 1. D 11. E. C 1. C 3. B 13. D 4. D 14. A 5. A 15. E 6. C 16. C 7. A 17. E 8. E 18. D 9. A 19. B 10. A 0. B Page 5
6 007 Physics Higher 1. (a) 1 cm: 50 m a = b + c bc.cos A 7 (drawn to scale) drawn to scale (1) (0) 60 ( ( ) 38 drawn to scale ) Displacement = 350 m at 038 / 38 o E of N / shown on correct diagram 38 as shown (on diagram) 350 m ±10 m 38 o ± o for correct diagram to scale, length and angle for adding correctly showing resultant direction = ( cos10º) a = 350 m a b sin A = sin B o = sin10 sin B B = 38 E of N 038( ) watch for 400 sin60 = 346m (0) o Can add rectangular components to get answer (b) v = s 350 = = 5. 3 ms -1 at 038 ( ) t 66 s or d acceptable Direction as (a) Or correct answer (038) (c) v = 6. 5 = s t 400 t 400 t = = (s) 6.5 (6) Car y arrives first. Symbol s or d acceptable No calculation (0) or Av. Speed of x: d v = t 400 = 66 1 = 6 1 ms ( ) Car y arrives first (d) 350 m at SW (0) 38 W of S -350m at m -350m at 038 or Consistent with (a) If measured from a correct diagram, accept 18 ± 1 Page 6
7 007 Physics Higher. (a) Component of weight = mgsinθ = sin(º) = 0 (N) use g = 10 or 9 81 deduct 00, 0 3 accept 0 7 deduct 1 7 (b) Unbalanced force = = 40 N (1) Must show unbalanced force F a = m 40 = 60 anywhere consistent with wrong (a), max (1½) a = 0/60 can get max = (m s - ) final line required - otherwise deduct (c) v = u + as = 0 + ( ) v = 8. m s -1 or E w = E k F u s = ½ mv = ½ 60 v v = 8 ms -1 (1) (1) (8 185) (8 19) (8) Max if a not 0 67 for all 3 formulae for all substitutions E p = mgh = = J E w = F.s = = 9000 J E k = 013 J = ½ mv v = 8. m s -1 (1) (d) Smaller mass smaller component of weight smaller unbalanced force smaller acceleration (not slower acc ) smaller speed at the bottom of the slope Must have smaller speed look for this first, otherwise (0) Force down slope/force parallel to slope not smaller weight + Speed less E p less, E w against friction, same, E k less But E p less, E k less, speed less 4 3 independent s Can do calculation with a smaller mass look for conclusion first if wrong (0) Page 7
8 007 Physics Higher Energy argument E k = E p Fd 1 m1 v 1 = m 1 gh Fd v 1 = gh Fd m 1 1 m v = m gh Fd v = gh m = Fd m larger smaller so v < v 1 v < v 1 Page 8
9 007 Physics Higher 3. (a) P 1 V 1 = P V = 15 V V = m 3 (1) (b) Volume of gas available = = m 3 (1) 0 48 no. = = (0) Number of balloons = but if followed by 0 08 in cylinder = 4 (1) = 0 (balloons) (1) so 4 4 = 0 (1) + (c) As mass of the helium gas is constant from the cylinder into the balloons, then as the volume increases, mass m and as density =, ρ = volume v then density must decrease in the balloons. OR As pressure has decreased, Number of collisions per second decreases Fewer molecules per unit volume mass density = 4 volume Density decreases Must have density decreases or (0) 3 statements are independent 4 density decreases as P = ρgh (0) if argued in converse, must then explicitly say that original density in cylinder is greater. + Page 9
10 007 Physics Higher 4. (a) At A, E k = ½ mv = ½ ( ) Must show squaring in nd line otherwise 6 = (J) Increase in E k = work done between the plates = = (J) final line required - otherwise deduct J W.P. so (1½) max (b) W = QV = V V = V (1) E (w) = QV acceptable 15 Must use J - otherwise max Deduct for 531 V 5313 V Acceptable sig. fig. 5310V, 5300V (c) (Same p.d.) But the charge is smaller independent less work is done smaller (increase in) kinetic energy (1) Look for conclusion first E k less, then. Smaller mass is irrelevant. Can be done by calculation but must have final statement + Page 10
11 007 Physics Higher 5. (a) E = 1 V (1) Deduct for no units 1 6 (b) (i) E = V + Ir 1 = (I. 0). 4 = I. 0 Lost volts = Ir (1 9. 6) = I I = 1. A (1) I = 4 0 = 1. A (1) or consistent with (a) v = IR V = IR 4 = I I = 1 A () 9 6 = I I = 4 8 A (0) (ii) R = V 9 6 = I 1 = 8 Ω (1) or consistent with b(i) If negative in final answers, deduct 1 Page 11
12 007 Physics Higher Question 5 (continued) (c) p.d./(v) 1 8 R ext = 4 Ω E = I ( R ext + r) 1 = I (4 + ) I = A V = IR (0) (S closed) time/(s) = x 4 shape and 1V or consistent with (a) (1) 8 calculation 8 on graph = 8 (V) Copy original shape and values - max + Partial marks: Must have at least labels on axes of V and t otherwise deduct Check graph if all values correct then () Check calculations go on to allow other than 8V only if consistent with (a) and (b) (ii) or arith error R = 4 (Ω) I = 1 (A) (0) W.P. stop V = 4 8V Could get (1) with no graph Page 1
13 007 Physics Higher 6 (a) I = V R 7 1 = Wrong power of 10 deduct = A (1) (b) V C = = 8. V (1) Q = CV independent = max if 3.8 or 1 used for V if to be treated as arith error, must show subtraction = C (1) (accept C) 3+ (c) E = ½ CV Prefix error: = ½ = J (1) (accept J) Once per prefix in each question ie value of R in (a) value of C in (b) or (c) Q = CV = = C E = ½ QV = ½ = J (1) both formulae both substitutions Page 13
14 007 Physics Higher 7. (a) X Y = Z RTh But R 1 /R = R 3 /R 4 acceptable = R Th 750 R Th = 330 Ω (1) (b) (i) Differential (mode) (1) Difference mode (0) 1 (ii) As temperature decreases: 1st statement voltage across decreases voltage at Q decreases voltage across thermistor increases potential at Q decreases potential difference between P and Q increases amplified (by the op amp) MOSFET/transistor switches on/conducts when voltage reaches threshold/certain/sufficiently high voltage But not at 0 7V 4 independent s instead of first (½ marks) Bridge out of balance causes p.d. ie if this ans then amplified by op amp MOSFET on (max 1½) Voltage flowing loses the each time + (iii) V o = Rf R1 (V V 1 ) V o = Rf R1 (V 1 V ) accept 3. 0 = 100 (V V 1 ) =. 5 V PQ V PQ = 1. V (1) Watch for inverting mode equation W.P. (0) Unless using it twice and subtracting answers to get difference V PQ = (V V 1 ) accept Page 14
15 007 Physics Higher 8. (a) maximum - constructive interference bigger crest and bigger trough bigger amplitude Waves must meet / combine / overlap 8 Waves meet - in phase / in step (1) or crest & crest and trough & trough or path difference is nλ Or by diagram + = 1 (b) (i) (A) Mean AB = = = m (1) Deduct if no unit 1 1m (1) 1m (0) outwith range 1 (B) Random uncertainty = = m (1) Do not deduct for no unit in both A and B 1 (ii) % AB = 0 01 ( 100) = 0. 9 % 1 10 Missing % deduct once % BC = 10 ( 100) = 3. 7 % 70 (BC has the larger percentage uncertainty) Must have percentage answers Page 15
16 007 Physics Higher Question 8 (b) continued (iii) nλ = d sinθ x λ = λ = (m) 3. 7% of = λ = ( ± 0. (1) ) m ± Uncertainty must be consistent decimal places or deduct. 4% of = m ± 3 7% (max 1½) (4. 9 ± 0. ) 10-7 m m ± m 3+ Page 16
17 007 Physics Higher 9. (a) sinθ 1 sinθ = n or sinθ a sinθ g sin 40 = 150 sinθ 5 sin 50 sinθ () o = θ = 5 4 o for formula o θ = 31 (1) o ( 30 7 ) deduct if degree sign missing in final answer (b) n = λ1 λ sin sinθ θ 1 = λ1 λ sin sinθ θ 1 λ = 1 λ = λ 1 40 sin 40 λ = 1 sin sin 50 λ1 = sin λ 1 = = 630 nm (1) λ 1 = 65nm () if 40 used also in (a) λ 1 = 65nm (1) or consistent with (a) (c) Blue light has a higher/larger frequency different frequency independent s due to a different refractive index. a larger refractive index refracts more refracts at greater/different angle not bends not different path (in question) 1+ Page 17
18 007 Physics Higher 30. (a) f = λ v = (= Hz) E = h f (o) independent = OR = (J) E = h λ v (1) final line required - otherwise deduct If: E = hf (0) = = (J) (1) = = J ie frequency has just appeared max for equation (b) (i) E k = x Negative answer W.P. (0) = J (1) 1 (ii) Current/Ammeter reading decreases (1) Look for A reading first. Irradiance decreases: fewer photons hitting plate per second fewer electrons released/one electron per photon Rate required for marks Max (1½) if this is not mentioned anywhere in the answer If go on after correct answer with W.P. then deduct I α I not justified Ignore intensity (the word) I = Nhf on own (0) Less radiation on plate (0) + Page 18
19 007 Physics Higher 31. (a) Decrease in mass = ( ) = (kg) E = mc independent If truncated mass values used, then max for E = mc. If don t show square value E = then max (1½) 6 = ( ) = J (1) (b) (D = D t) = = (Gy) H = D w R anywhere = = Sv (1) = (4 µsv) or H = D W R = 4 3 = 1 µsvh -1 H = H t 1 = H H = 4 µsv (1) E = mc before and after and then subtract energies without rounding can get (3) All ( 10-6 ) terms may be omitted and final answer given in µsv. H = D.W R anywhere H = D.W R (½ max) = (WP) Can change to seconds, but if only change one quantity then (½) max - gives H = Sv Accept Q W Instead of W R 3 3 [END OF MARKING INSTRUCTIONS] Page 19
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