2005 Physics. Advanced Higher. Finalised Marking Instructions

Size: px

Start display at page:

Download "2005 Physics. Advanced Higher. Finalised Marking Instructions"

Rodger Montgomery
6 years ago
Views:

1 005 Physics Advanced Higher Finalised Marking Instructions These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments.

2 Detailed Marking Instructions AH Physics 005. Numerical Marking (a) (b) (c) (d) (e) 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. Negative marks or marks to be subtracted should not be shown. An inverted vee may be used instead. 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. Marks awarded should be transferred to the script booklet inner margin and marked G. Fractional marks, if awarded to individual questions, should be recorded in the grid, but the total script mark must be rounded up to the next whole number when transferred to the box at the top of the script.. 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. 3. 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. General Instructions (Refer to National Qualifications Booklet) (a) (b) (c) (d) (e) (f) (g) (h) 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. 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. 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. (i) (j) (k) (l) 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. 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 higher and one lower Page 3

4 (m) Squaring Error E K = ½ mv = ½ 4 = 4J (-½, ARITH) E K = ½ mv = ½ 4 = 4J (½, formula) Incorrect substitution. The General Marking Instructions booklet should be brought to the markers' meeting. Page 4

5 Physics Marking 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.. V=IR 7 5= 5R R=5 0Ω Answers Mark +comment Issue () Ideal Answer. 5 0Ω () Correct Answer GMI Unit missing GMI (a) Ω (0) No evidence/wrong Answer GMI 5. Ω (0) No final answer GMI V 7 5 Arithmetic error GMI 7 R= = =4 0Ω I 5 R= I V =4 0Ω Formula only GMI 4 and R= I V = Ω Formula only GMI 4 and V 7 5 () Formula + subs/no final answer GMI 4 and R= = = Ω I 5 V 7 5 () Formula + substitution GMI (a) and 7 R= = =4 0 I 5 V 5 Formula but wrong substitution GMI 5 R= = =5 0Ω I 7 5 V 75 Formula but wrong substitution GMI 5 R= = =5 0Ω I 5 I 7 5 (0) Wrong formula GMI 5 R= = =5 0Ω V 5 4. V=IR 7 5= 5 x R R=0 Ω Arithmetic error GMI 7 5. V=IR I 5 Formula only GMI 0 R= = =0 Ω V 7 5 Page 5

6 Data Sheet Common physical Quantities Quantity Symbol Value Quantity Symbol Value Gravitational acceleration on Earth Radius of Earth Mass of Earth Mass of Moon Mean of Radius of Moon Orbit Universal constant of gravitation Speed of light in vacuum Speed of sound in air g R E M E M M G c v 9 8 Ms 6 4 x 0 6 m 6 0 x 0 4 kg 7 3 x 0 kg 3 84 x 0 8 m 6 67 x 0 - m 3 kg - s x 0 8 ms x 0 ms - Mass of electron Charge on electron Mass of neutron Mass of proton Mass of alpha particle Charge on alpha particles Planck s constant Permittivity of free space Permeability of free space M e e m n m p m a h ε 0 µ 0 9 x 0-3 kg - 60 x 0-9 C 675 x 0-7 kg 673 x 0-7 kg 645 x 0-7 kg 3 0 x 0-9 C 6 63 x 0-34 Js 8 85 x 0 - Hm - 4π x 0-7 Hm - Refractive Indices 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 Aluminium Copper Glass Ice Gylcerol Methanol Sea Water Water Air Hydrogen Nitrogen Oxygen Density/ Kg m x x x x 0 6 x x 0 0 x x x Melting Point/K Boiling Point/K Specific Heat Capacity/ Jkg - K x x x 0 0 x x x x x x x x 0 Specific Latent Heat of Fusion/ Jkg x x x x x x Specific latent Heat of Vaporisation/ Jkg 8 30 x 0 5 x x x x x 0 5 The gas densities refer to a temperature of 73 K and pressure of 0 x 0 5 Pa. Page 7

8 005 AH Physics.(a) ω = v 9 r = =56 5 rad s - () (b) ω = v r = = ( 4 rad s - ) (c) ωr is constant r increases (ω decreases) Accept v = r ω (d) (i) θ = no. of revolutions π = () = ( radians) (d) (ii) ω = ω 0 + αθ 4 = α α = max = rad s () (d) (iii) α = ω - ω o t t = ω - ω o α = = 4460 s () ( 74 3 minutes) Accept using θ = ω 0 t + ½ αt ω 0 <ω - max for equation Page 8

9 005 AH Physics (a) I child = mr = 5 = 00 (kg m ) I total = I roundabout + I child = for adding ( = 600 kg m ) (b) The angular momentum before (an impact) equals the angular momentum after the impact provided there are no external torques. (c)(i) mv = 5 4 = 60 kg m s - () (c)(ii) mrv = 5 4 = 0 kg m s - () (d) I ω = I ω 0 = 600 ω ω = = 0 rad s - () (e) E k before = ½ I ω = = 7 (J) E k after = ½ I ω (formula) = = (J) E k lost = 7 - = 60 J Isolated system acceptable or ω = v/r = 4/ = (rad s - ) Iω = 00 = 0 kg m s - () or ½ mv = = 7 (J) (f) ω = ω 0 + αθ 0 = 0 + α (0 5 π) α = ( rad s - ) T = I α = 600 (-) = (-) 3 8 N m () Or E K = Tθ () = Tπ T = π = 3.8Nm () 3 Page 9

10 005 AH Physics 3(a)(i) mω R = GMm () 8 R ω = GM R 3 cancelling m ( π/t) = GM R for ω = π/t T = 4π R 3 GM (a)(ii) T = 4π R 3 GM = ( ) 3 data data T = = s () GM (b)(i) E p = m R = substitution ( ) () for adding (= J) (b)(ii) E k = ½mv = ( ) = J E total = E p + E k = = J () E T = -GMm R = ( ) = J () Page 0

11 005 AH Physics 4(a) Acceleration is proportional to displacement (from a Accept F = - k x 7 fixed point) or a = - k x and is always directed to (that) fixed point. or The unbalanced force is proportional to the displacement (from a fixed point) and is always directed to (that) fixed point (b)(i) (x = Asinωt) v = -ωacosωt ω = 65 (rad s - ) f = ω π = = 99 5 Hz () (00 Hz) (b)(ii) ωa = 0 5 A = = m () (c) (maximum) acceleration = 9 8 m s - () (contact lost) when cap accelerates downwards greater (or equal to) g or similar () Must have 9 8 ms - to gain nd mark or 0 cone s acceleration (or speed) greater than g or greater than bead s acceleration (or bead) Page

12 005 AH Physics 5(a) E = Q Q formula for E 7 + 4πε 0 r 4πε 0 r = π (3 0-3 ) 4π ( 0-3 ) substitution must have "+" = = N C - () (a)(ii) to the right () (b)(i)(a) () correct shape of lines to cylinder essential for the other marks direction outside lines straight or bulging (b)(i)(b) () or 0 (b)(ii) (external) electric fields (interference) cannot reach the central wire () idea of shielding E inside of mesh = 0 Page

13 005 AH Physics 6(a) QV = ½mv = v v = (v = m s - ) (b) t = d v (c)(i) = = s () V E = d = = 000 (N C - ) F = EQ = (= N) (c)(ii) a = F m 3 = = 0 5 (m s - ) s = ut + ½ at = ( ) = m () Page 3

14 005 AH Physics 6 (d)(i) There is an unbalanced force (on the electron) () Accept vertical acceleration in the vertical direction () or Electrons attracted to positive plate (d) (ii) No (unbalanced) forces act (on the electron) () (e) s increases () since v decreases t decreases Page 4

15 005 AH Physics 7(a)(i) positive () 0 (a)(ii) Bqv = (½ ) mv R = = C kg - () Accept m = 04 x 0-8 kg c - q q = 9 6 x 0 7 m no unit 3 (a)(iii) proton () q = () m = (C kg - ) (same) (b) the component of the electron's velocity perpendicular to the magnetic field causes circular motion () (or equivalent) the component of the electron's velocity parallel to the magnetic field is unchanged () (or no force on electron parallel to B) Good answer () Some valid physics in description () (c) Enter toward the poles () move in circles/spirals () or never reach atmosphere above equator () Page 5

16 005 AH Physics 8(a) volts is induced in the coil when the current 9 changes at (a rate of) A s -. () (b) E = - L di dt - = - di dt minus sign missing 0 marks (E is back emf across L) di = - max (formula) dt di = dt = 6 A s - () (c)(i) I max is less due to V s less calculation evidence I max = 0 4 = 5 (A) Allow justification with no calculations (initial) di is greater dt calculation evidence di = 0 dt 5 since L is smaller = 6 7 (A s - ) (c)(ii) E = ½LI = = 4 7 J () Page 6

17 005 AH Physics 8(d)(i) (gold bracelet) moves in magnetic field () (conductor) (d)(ii) moving magnetic field or changing current induces current (voltage) Page 7

18 005 AH Physics 9(a)(i) πf = 570 () 9 f = = 50 Hz () (a)(ii) y = sin(570t + 4 6x) () () for amplitude for plus sign (b)(i) frequency increases approaching and decreases after train passes (b)(ii) The waves (wavefronts) are closer together as they approach the person () then they are further apart after they pass the person () or Any statement of speed of sound increasing or decreasing (0) () () (b)(iii) f = f s v v + v s 760 = v s v s = 358 v s = 8 m s - () "-" used 0 marks "±" used does not get formula unless "+" is selected in the next line Page 8

19 005 AH Physics 0(a)(i) λ = xd 9 D = = m () (a)(ii) % uncertainty in x = = 6 5% 8 % uncertainty in d = = 4% 0 5 ( % uncertainty in D ) = = 0 3% ) 3 9 % uncertainty in λ = = 7 4 % Allow 6% or absolute uncertainty in λ = 7 4% = m (a)(iii) An uncertainty should be quoted to one significant figure Too many significant figures okay Too many decimal places (0) (b) (new) % uncertainty in x = = 0 8% 64 (new) % uncertainty in λ = 4% new absolute uncertainty in λ = 4% = 0-8 m () (c)(i) The % uncertainty in x is very small () compared to the % uncertainty in d or reducing 0 8% still further does not change the uncertainty of λ at 4%. or (c)(ii) the slit separation (d) () Page 9

20 005 AH Physics (a) Unpolarised light has the (electric field) oscillating 6 in all planes. Accept diagrams Polarised light has the electric field oscillating in one plane only. () Plus explanation In all directions (0) In one direction (0) (b)(i) (Polarised) light cannot pass through the liquid crystal () and is not reflected by the mirror. () (b)(ii) Switch is opened. () (c) The numbers disappear (or cannot be seen) and then re-appear. () Light reflected from calculator polarised Indication of polarising material blocking/allowing transmission of light depending on rotation. [END OF MARKING INSTRUCTIONS] Page 0

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