2012 Physics. Advanced Higher. Finalised Marking Instructions

Transcription

1 01 Physics Advanced Higher Finalised ing Instructions Scottish Qualifications Authority 01 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 Delivery: Exam Operations. 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 Delivery: Exam Operations 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. 1. 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 1 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 1 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) (m) Significant figures. Data in question is given to 3 significant figures. Correct final answer is 8 16J. Final answer 8 J or 8 158J or J No penalty. Final answer 8J or J 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. Max ½ mark deduction per question. Max ½ deduction from question paper. Squaring Error E K = ½ mv = ½ 4 = 4J Award 1½ 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 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 50 Answers + comment Issue Ideal Answer. 5 0Ω () Correct Answer GMI (1½) Unit missing GMI (a) (0) No evidence/wrong Answer GMI 1 5. (0) No final answer GMI V 7 5 (1½) Arithmetic error GMI 7 R= = =4 0 I 1 5 R= I V =4 0 Formula only GMI 4 and 1 R= I V = Formula only GMI 4 and 1 V 7 5 Formula + subs/no final answer GMI 4 and 1 R= = = I 1 5 V 7 5 Formula + substitution GMI (a) and 7 R= = =4 0 I 1 5 V 1 5 Formula but wrong substitution GMI 5 R= = =5 0 I 7 5 V 75 Formula but wrong substitution GMI 5 R= = =5 0 I 1 5 I 7 5 (0) Wrong formula GMI 5 R= = =5 0 V V=IR 7 5=1 5 R R=0 (1½) Arithmetic error GMI V=IR I 1 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 1 s Planck s constant h Js Speed of light in vacuum c ms 1 Permittivity of free space ε Fm 1 Speed of sound in air v ms 1 Permeability of free space µ Hm 1 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 Density/ kg m -3 Melting Point/K Boiling Point/K Specific Heat Capacity/ Jkg -1 K -1 Specific Latent Heat of Fusion/ Jkg -1 Specific latent Heat of Vaporisation/ Jkg -1 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 Section A Question Expected Answer/s Max 1 a i m m 0 v 1 c Look out for correctly calculating m. m= = kg v 1 (310 8 ) 8 1 v.510 ms 1 a ii E mc Must have correct mass 31 8 E ( ) 13 E J 1 a iii Weak (force) 1 1 b i Electron Diffraction Interference Fire electrons through crystals Thomson-Reid Experiment 1 b ii Compton Effect Photoelectric effect e/m experiment electrons deflected in a deflection tube electron s back scattering 1 Young s/ Double Slit Bending (0) If explanation contradicts the example then WP (0) Defraction (0) 1 Electrons repel Any indication of Force due to electrostatic effects Electrons can be accelerated Electrons have mass/charge / momentum (0) Nuclear Fission (0) If explanation contradicts the example then WP (0) Milikan s oil drop (0) Rutherford s expt (0) Defract (0) Must describe an effect Page 8

9 a i dv a dt dv a. dt or dv dt dt a. dt dv a dt v dv u t 0 a. dt + v at c ½ for integrals, ½ for limits need both before can progress at t = 0, c = u v t [ ] [ at] Must be specific with respect to time v u at SHOW ME v u 0 v u at (0) v u at a ii v ( u at)( u at) 1 SHOW ME Starting with s=ut+ ½at ½ for substitution for t ½ for manipulation v u uat a t v 1 u a( ut at ) Check second line both a and t are squared. v u as SHOW ME b i s m v u as a 14.9 a 3.1 ms Page 9

10 b ii v u as 3 v r a ( 8.0) + sub and value of s o a 0.5 ms -1 o 1.18 rad s a r for both substitutions for and o rad s 0.06 rad s o rad s c i An indication of the central/ inward force independent marks Provided by (horizontal) component of F Any indication of outwards/centrifugal force (0) Sideways no indications of central (0) c ii (Central) force is no longer large enough to maintain her circular motion 1 Any indication of outwards/centrifugal force (0) The centripetal force is now greater than the frictional force Page 10

11 3 a i y A sint or ω πf 0.76 y A cos t Accept π in final answer -1 ω 48. (rad s ) ω=1.5π A 0.18 (m) y 0.18 sin4. 8t Or y 0.18 cos4 8t 3 a ii g ( ) A 1 Must start with the equation 9.8 A 4.8 a ( ) A Value of g must appear A 0.43 m SHOW ME OK to calculate a using A= b i Assume diver rods about one end 33.0 kg per rod 3 NB 3 mark question Accept r instead of l in equations Acceptable to use 1 I ml 1 I ml 3 1 I (kg m ) 3 I I (kg m ) I 18.6 kg m 3 I 18.6 kg m Cannot use an average length 3 b ii Some indication of uneven mass distribution. 1 Diver not rigid (0). Distribution of weight (0). Consideration of uneven length(0) 3 b iii ( L ) I I ω -1 ω 0.74 rad s Page 11

12 3 c i 1 E krot I NB Only marks I E krot J Accept 1.6 J for either J Accept.1 J 1 1 krot E 0.54 J Depending on rounding can be 0.57 to 0.55J Accept if negative change in E krot 3 c ii Work is being done by the diver 1 Energy provided by diver or equivalent Diver pulls his legs in Page 1

13 4 a GM m 1 p F mg both equations g r m for equating GMm SHOW ME g mr 4 b i GM g r Accept 7.06 or 7.00 from graph 6 r m V (1. 10 ) g 6 (1. 10 ) 0.59 N kg -1 g or ms V g V R p g g 0.58 N kg If not double negative then WP ½ for both formula ½ for both substitutions 1 for final answer 4 b ii GMm E r No negative in equation (0) No negative in sub max No negative in answer (1½) max Or from the graph accept values of V from -4.7 to E J E Range 7 E J 7 E 5.6 to J V GMm r (0) Page 13

14 4 c x 3 If subscripts on M s and r s then can get for equating two forces. Ignore loose subscripts on masses if denominators OK GM x p m GM m c 7 ( x ) 7 1 M M p c or M C = ( marks if x+ y= defined) Ensure masses are above the correct denominator 7 1 x 7 ( x) 7 x m from Pluto 7 x m from Pluto Page 14

15 5 a Electrons/negative charges in sphere move to rhs of sphere (leaving +ve charge on lhs of sphere). 1 Diagram Must have roughly equal numbers of + charge and charges for mark Any indication of sphere charged (0). Movement of protons /positively charged particles (0) Movement of positive charges 5 b i 3 3 V (.0 10 ) V If V= 1 kv then max V E V / d / V m or NC ( V m ) If use 5.0 V then (1½) max 5 b ii F qe Ignore negatives N N possible if number carried over in calculator. Page 15

16 5 b iii tan F mg F T sin sin3.4 mg or T cos as an alternative eq value of mg eq 3 T T cosθ θ T sinθ mg Fe T Or N (0.047) F T ( ) (3.910 ) T N tan T mg for Pythag and for value of mg Treat and magnitude as independent 3. 4 () marks for calculating T mark for correct angle. 5 c Angle is unchanged Uniform electric field/force acting is constant Angle increases/decreases (0) Page 16

17 6 a i (Point) Q 1 6 a ii mv Bqv r q m v rb 1 SHOW ME Deduct for any subsequent mistakes 6 a iii q / m v rb Calculation () Statement (Ckg ) Calculations independent Alpha particleq / m Do not penalise for wrong unit (Ckg ) Particle is alpha Justification needed for this mark 6 a iv t d / v Calculation of T t=t/ r / v 6.910, both Both substitutions s 8 If s rounded t.610 s 6 a v t is constant As v doubles, r doubles t d / v r 1 / v1 r1 / v1 both v and r double or directly proportional as before Not enough to say r increases 6 b i Particle is negatively charged 1 Smaller charge (0) 6 b ii Charge to mass ratio is smaller 1 Must be q/m not m/q Page 17

18 7 a i h p m NB 3 sig fig is correct so is max (1½) 7 a ii If an e.m.f. of ±0.1 V is induced when the current is changing at the rate of 1 A s -1, the inductance is 0.1 H. 1 Or equivalent, eg -1 1 V for 10A s 7 b i Magnetic field strength increases and reaches a maximum value/levels off 1 Any mention of magnetic field strength decreasing/ changing (0) 7 b ii At -1 t 0, di / dt 4.0 A s. At this time E 1V Accept L=1/4 assume cancelled E L di / dt 1 L 4 L 3.0 H 7 b iii V/V If general trend of graph wrong (0) V Labels= 0, , t(s), V(V) shape Value of V missing or incorrect maximum mark all labels etc Must have origin for labelled for labels marked. 0 t/s Accept 1.6 s Maximum current E / Rt 1/101. A Or use voltage divider to find V Maximum p.d. across 8 8 V V V Page 18

19 7 b iv 1 E LI Look out for carry forward of wrong answer Must have current = 1.A, cannot. J carry through wrong current 7 c Reading on A 1 will increase as (capacitive) reactance decreases/ If Reading on A will decrease as inductive reactance increases/ I1/f Alternative answer A 1 at higher frequencies the current drop in each half cycle is less. A back emf increases Impedence OK Resistance not OK Page 19

20 8 a 0I B r T 8 b i (The current in each wire produces) a magnetic field. Same direction of the magnetic fields between the wires. OR interpretation of F=BIl May use diagrams. x Diagram Explanation 8 b ii F l 0I1I r I r I SHOW ME r 1 SHOW ME Must show µ o =4π 10-7 or b iii Weight per unit length (N m ) If there is no calculation or value for weight (0) I I 67A Page 0

21 9 a i Read from the graph λ 0.5m 1 No tolerance 9 a ii Read from the graph A 0.58m 1 accept 0.57 to 0.585m 0.6m (0) 9 a iii v fλ 1 1 mark for the answer 1.5 f 0.5 Can be carry through from 9ai f 5.0 Hz 9 a iv Or 3/ 4 rad or 3π/ No tolerance in reading from graph accept 4.77 or 4.78 or 4.8 Phase angle (rad) Answer not required but incorrect unit (-½) e.g. rads Can be carry through from 9ai 9 b x y ( )0.58sin (5.0t ) 0.5 () If not travelling wave equation max for A and Accept y ( )0.58sin(31t 5x) y ( )0.58sin (5.0t 4.0x) y ( )0.58sin(10.0 t 8.0 x) for A, for t term, for x term, for negative sign 9 c y ( )0.9sin(31t 5x) 1 If not travelling wave equation max for A No requirement for same or f as part 9b. x y ( )0.9sin (5.0t ) 0.5 for 0.9, for positive sign Accept 0.3 but not 0.30 Page 1

22 Question Expected Answer/s Max 10 a i v ds/ dt both equations mark 8.t m s m s 10 a ii Frequency is increasing/increases Waves become more and more squashed together as speed increases or time between wave continually decreasing higher frequency (0) Frequency has increased (0) Can link to Doppler equation with v s increasing 10 b f v fs v vs Hz f 65 Hz with v If carry through of 8. m s -1 in part 10ai then f = 610 Hz Page

23 11 a i x D / d NB mark question Gradient of graph If gradient only = 0.06 (0) ( ) Accept values from (3 6) 10 Accept correct use of 1/gradient 58.8 gradient Do not penalise units in gradient at this stage. 9 d d m d can = m 11 a ii Uncertainty too small 1 Very precise measurement 11 a iii Any two from Measure distance between several spots Use a bigger range of values for D Increase value(s) of D Additional data points (mark spots on paper) and use travelling microscope to find x each Repeated measurements (0/1) max mark for other correct statement Do not accept measure from middle of spots. Take more readings (0) Must specify higher precision instrument increase in (0/1) decrease d (0/1) 11 b Spots blurred/ elongated in horizontal direction Accept more elongated in the vertical direction Spacing increased Accept dimmer [END OF MARKING INSTRUCTIONS] Page 3