2014 Physics. Advanced Higher. Finalised Marking Instructions

Size: px

Start display at page:

Download "2014 Physics. Advanced Higher. Finalised Marking Instructions"

Antony Lloyd
5 years ago
Views:

1 04 Physics Advanced Higher Finalised ing Instructions Scottish Qualifications Authority 04 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial basis. If it is to be used for any other purposes written permission must be obtained from SQA s NQ Assessment team. 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 NQ Assessment team may be able to direct you to the secondary sources. These ing Instructions have been prepared by Examination Teams for use by SQA Appointed ers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

2 Part One: General ing Principles for Physics Advanced Higher This information is provided to help you understand the general principles you must apply when marking candidate responses to questions in this Paper. These principles must be read in conjunction with the specific ing Instructions for each question. (a) s for each candidate response must always be assigned in line with these general marking principles and the specific ing Instructions for the relevant question. If a specific candidate response does not seem to be covered by either the principles or detailed ing Instructions, and you are uncertain how to assess it, you must seek guidance from your Team Leader/Principal Assessor. GENERAL MARKING ADVICE: Physics Advanced Higher The marking schemes are written to assist in determining the minimal acceptable answer rather than listing every possible correct and incorrect answer. The following notes are offered to support ers in making judgements on candidates evidence, and apply to marking both end of unit assessments and course assessments.. Numerical ing (a) (b) (c) (d) The fine divisions of marks shown in the marking scheme may be recorded within the body of the script beside the candidate s answer. If such marks are shown they must total to the mark in the inner margin. The number recorded should always be the marks being awarded. The number out of which a mark is scored SHOULD NEVER BE SHOWN AS A DENOMINATOR. (½ mark will always mean one half mark and never out of.) Where square ruled paper is enclosed inside answer books it should be clearly indicated that this item has been considered. s awarded should be transferred to the script booklet inner margin and marked G. The total for the paper should be rounded up to the nearest whole number.. Other ing 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.) Excess significant figures. 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. X Wrong Physics Wrong order of marks No other annotations are allowed on the scripts. Page

3 3. General Instructions (Refer to National Qualifications ing Instructions Booklet) (a) (b) (c) No marks are allowed for a description of the wrong experiment or one which would not work. Full marks should be given for information conveyed correctly by a sketch. Surplus answers: where a number of reasons, examples etc are asked for and a candidate gives more than the required number then wrong answers may be treated as negative and cancel out part of the previous answer. Full marks should be given for a correct answer to a numerical problem even if the steps are not shown explicitly. The part marks shown in the scheme are for use in marking partially correct answers. However, when the numerical answer is given or a derivation of a formula is required every step must be shown explicitly. (d) (e) (f) (g) (h) Where mark is shown for the final answer to a numerical problem ½ mark may be deducted for an incorrect unit. Where a final answer to a numerical problem is given in the form 3 6 instead of then deduct ½ mark. Deduct ½ mark if an answer is wrong because of an arithmetic slip. No marks should be awarded in a part question after the application of a wrong physics principle (wrong formula, wrong substitution) unless specifically allowed for in the marking scheme eg marks can be awarded for data retrieval. In certain situations, a wrong answer to a part of a question can be carried forward within that part of the question. This would incur no further penalty provided that it is used correctly. Such situations are indicated by a horizontal dotted line in the marking instructions. Wrong answers can always be carried forward to the next part of a question, over a solid line without penalty. The exceptions to this are: where the numerical answer is given where the required equation is given. (i) (j) (k) ½ mark should be awarded for selecting a formula. Where a triangle type relationship is written down and then not used or used incorrectly then any partial ½ mark for a formula should not be awarded. In numerical calculations, if the correct answer is given then converted wrongly in the last line to another multiple/submultiple of the correct unit then deduct ½ mark. Page 3

4 (l) Significant figures. Data in question is given to 3 significant figures. Correct final answer is 8 6J. Final answer 8 J or 8 58J or 8 576J No penalty. Final answer 8J or 8 576J Deduct ½ mark. Candidates should be penalised for a final answer that includes: three or more figures too many or two or more figures too few. ie accept two more and one fewer. (m) Squaring Error E K = ½ mv = ½ 4 = 4J Award ½ Arith error E K = ½ mv = ½ 4 = 4J Award ½ for formula. Incorrect substitution. The General ing Instructions booklet should be brought to the markers meeting. Page 4

5 Physics ing Issues The current in a resistor is 5 amperes when the potential difference across it is 7 5 volts. Calculate the resistance of the resistor. Answers + comment Issue. V=IR Ideal Answer 7 5= 5R R 50 (). 5 0Ω () Correct Answer GMI (½) Unit missing GMI (a) (0) No evidence/wrong Answer GMI 5. (0) No final answer GMI V 7 5 (½) Arithmetic error GMI 7 R= = =4 0 I 5 R= I V =4 0 Formula only GMI 4 and R= I V = Formula only GMI 4 and V 7 5 () Formula + subs/no final R= = = I 5 answer GMI 4 and V 7 5 () Formula + substitution GMI (a) and 7 R= = =4 0 I 5 V 5 Formula but wrong R= = =5 0 I 7 5 substitution V 75 Formula but wrong R= = =5 0 I 5 substitution GMI 5 GMI 5 I 7 5 (0) Wrong formula GMI 5 R= = =5 0 V 5 4. V=IR 7 5= 5 R R=0 5. V=IR (½) Arithmetic error GMI 7 I 5 Formula only GMI 0 R= = =0 V 7 5 Page 5

6 DATA SHEET COMMON PHYSICAL QUANTITIES Quantity Symbol Value Quantity Symbol Value Gravitational acceleration on Earth g 9 8 ms Mass of electron m e kg Radius of Earth R E m Charge on e C electron Mass of Earth M E kg Mass of neutron m n kg Mass of Moon M M kg Mass of proton m p kg Radius of Moon R M m Mass of alpha particle m kg Mean Radius of Moon Orbit m Charge on alpha particle C Universal constant of gravitation G m 3 kg s Planck s constant h Js Speed of light in vacuum c ms Permittivity of free space ε Fm Speed of sound in air v ms Permeability of free space µ Hm REFRACTIVE INDICIES The refractive indices refer to sodium light of wavelength 589 nm and to substances at a temperature of 73 K. Substance Refractive index Substance Refractive index Diamond Glass Ice Perspex Glycerol Water Air Magnesium Fluoride SPECTRAL LINES Element Wavelength/nm Colour Element Wavelength/nm Colour Hydrogen Red Blue-green Blue-violet Violet Ultraviolet Ultraviolet Cadmium Red Green Blue Lasers Element Wavelength/nm Colour Sodium 589 Yellow Carbon dioxide Helium-neon Infrared Red Page 6

7 PROPERTIES OF SELECTED MATERIALS Substance Density/ kg m -3 Melting Point/K Boiling Point/K Specific Heat Capacity/ J kg - K - Specific Latent Heat of Fusion/ J kg - Specific Latent Heat of Vaporisation /J kg - Aluminium Copper Glass Ice Gylcerol Methanol Sea Water Water Air Hydrogen Nitrogen Oxygen The gas densities refer to a temperature of 73 K and pressure of Pa. Page 7

8 Part Two: ing Instructions for each Question a i v r v 8 8 = 7 8 = rad s - () a ii Erot I 00 0 I 6 I = kgm () a iii There are no other moving parts in the blade system (e.g. gears). Accept Length of blade is the effective radius or blades are rigid/do not bend. b o t T = 0 06 IT = = Nm () (8) Page 8

9 a I mr I I = 3 0 kg m 5 () b v r = 0 (rads - ) mgh = mv I () I (sub) I I = 0 kg m 5 () c energy is lost or calculation assumes no energy is lost (7) Page 9

10 3 a i The (minimum) velocity/speed that a mass must have to escape the gravitational field (of a planet). 3 a ii E k + E p = 0 Therefore mv GMm 0 () r GM v r GM v r 3 a iii GM v r 3 v = () = m s - () 3 b = ms - () 6 Nitrogen, Oxygen, Methane, Carbon Dioxide could all be found on planet. () (8) Page 0

11 4 a The unbalanced force/ acceleration is proportional to the displacement of the object and act in the opposite direction. 4 b 007 m 4 c i t 500 = 60 3 = 57 (rad s - ) a y = ( ) = ( ) m s - () 4 c ii or Ek m A y Ek m A = = 85 J () (7) Page

12 5 a i Electrons behave like waves 5 a ii Photoelectric effect or Compton scattering Collision and transfer of energy 5 b h h or p mv m () 5 c h p = 0-34 m This value is so small (that no diffraction would be seen). Or the de Broglie wavelength of the bullet is much smaller than the gap. () 5 d i Electron orbits a nucleus / proton () Angular momentum quantised () or Certain allowed orbits / discrete energy level 5 d ii nh mvr = = kg m s - () (0) Alternative acceptable units Js kg m rad s - kg m s - rad - Page

13 6 a i QQ F PA 4 r ( N) = 0 05 N () 6 a ii (F RA ) = ( ) 0 0 (N) (F SA ) = ( ) 0 05 (N) 3 F QA = 0 0 N F (due to P&S) = 0 05 cos 37 = 0 04N (to right) F (due to Q&R) = 0 0 cos 37 = 0 09N (to right) Combined force = = N to right () ( N to right) F SA = 0 05 N Combine = 0 07 N F PA = 0 05 N F RA = 0 0 N Combine = 0 07 N 007N () 007N () 0043N () a = b + c bc cos A a = (0 07) + (0 07) ( cos 06 o ) a = N to right or (scale) drawing () Also accept sine rule (-½) if direction not given Page 3

14 6 b Q V= 4 r r r = 0 63 m () 6 b ii The work done (energy) used would be the same or The charge is in a conservative field, the path taken between two points does not affect the work done (energy used). (8) Page 4

15 7 a i qv = mv () (-½) if no final answer given q v. m V 7 a ii (A) q v m V Ckg 730 Ckg 700 Ckg 740 Ckg 640 Ckg 7 a ii (B) Ckg 5 7 a ii (C) Drawing a graph of v vs V () Draw graph of v against V () and calculating the gradient = () calculating the gradient as = () Page 5

16 7 b i QV = mv v v = ms - ( ½) if no final answer given or wrong final answer 7 b ii d t v t t = (s) QE = ma 900 E a = (m s - ) s ut at a s = 0 + ( ) s = m () 7 b iii Lower than x /vertical displacement reduces () Increased horizontal velocity Time between plates reduced () Page 6

17 8 a i qvb = mv r r = mv qb () ( ½) if no final answer given 8 a ii r = mv qb 05 r = = 0 5 m B B = 0 T () Page 7

18 8 a iii r T v m T = acceptable qb T= T = s () s for last line, wrong rounding using this method. 8 b i Accept any of the following πr T= v sin Period independent of velocity. Radius and vertical velocity reduce (in proportion) or both reduce Angular velocity is constant The magnetic induction has not changed mvsin r = so qb πmv sin T= qbv sin () m T = qb 8 b ii v vcos h v h cos 40 7 v h 84 0 m s pitch = v t pitch = pitch = m (8) Page 8

19 9 a i Out of page () 9 a ii To ensure that the accelerating potential is in the correct direction for the particle s motion. () Direction of force acting on charges reversed. 9 b i Ek mv = J 750 Number of gap transits b ii Their mass will increase (become relativistic) () A greater centripetal force will be required () or To keep the radius of orbit within the dimensions of the cyclotron qvb = mv r Must have increasing mass before second mark awarded. f = qb m To maintain a constant frequency, as their mass increases 9 c i A has no charge () B & C have different charge to mass ratios or B and C have opposite charges () 9 c ii The particles are losing energy or speed or momentum is decreasing (9) Page 9

20 0 a V I R I 48 Current (A) I = 0 5 (A) time () for shape labels 0 b di E = L dt di = dt di dt 3 0As () 0 c X L /(inductive reactance) increases Or back emf increases Therefore current decreases (impedance increases) (5) Page 0

21 a Coloured fringes are produced by interference () Reference to different colours produced by angle of viewing thickness of film optical path difference () b d 4n 5550 d = = m () c Wavelengths in the middle of the visible spectrum not reflected or destructively interfere. () Red and blue reflected / combined to (form purple). () (6) Page

22 a Division of wavefront b i D x d x x = 43 0 m () b ii % Uncertainty in λ = % 50 3 % Uncertainty in D = % 5 % Uncertainty in d = % % Uncertainty in x = % Absolute uncertainty in 3 x = 39% 43 0 = m () b iii Slit separation Highest percentage uncertainty (7) Page

23 3 a i The tablet emits plane polarised light. 3 a ii The brightness would gradually reduce from a maximum at 0 degrees to no intensity at 90 degrees. () It would then gradually increase in intensity from 90 degrees to 80 where it would again be at a maximum. () 3 b tan θ = n θ = Brewsters angle tan θ = 33 θ = 53 o θ = = 36 9 o (5) [END OF MARKING INSTRUCTIONS] Page 3

2006 Physics. Advanced Higher. Finalised Marking Instructions

2006 Physics. Advanced Higher. Finalised Marking Instructions 006 Physics Advanced Higher Finalised Marking Instructions The Scottish Qualifications Authority 006 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial