GCE A LEVEL MARKING SCHEME

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1 GCE A LEVEL MARKING SCHEME SUMMER 07 A LEVEL (NEW) PHYSICS - COMPONENT A40U0-

2 INTRODUCTION This marking scheme was used by WJEC for the 07 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was taken so that reference could be made to the full range of candidates' responses, with photocopied scripts forming the basis of discussion. The aim of the conference was to ensure that the marking scheme was interpreted and applied in the same way by all examiners. It is hoped that this information will be of assistance to centres but it is recognised at the same time that, without the benefit of participation in the examiners' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

3 MARK SCHEME GENERAL INSTRUCTIONS Recording of marks Examiners must mark in red ink. One tick must equate to one mark (except for the extended response question). Question totals should be written in the box at the end of the question. Question totals should be entered onto the grid on the front cover and these should be added to give the script total for each candidate. Marking rules All work should be seen to have been marked. Marking schemes will indicate when explicit working is deemed to be a necessary part of a correct answer. Crossed out responses not replaced should be marked. Credit will be given for correct and relevant alternative responses which are not recorded in the mark scheme. Extended response question A level of response mark scheme is used. Before applying the mark scheme please read through the whole answer from start to finish. Firstly, decide which level descriptor matches best with the candidate s response: remember that you should be considering the overall quality of the response. Then decide which mark to award within the level. Award the higher mark in the level if there is a good match with both the content statements and the communication statement. Marking abbreviations The following may be used in marking schemes or in the marking of scripts to indicate reasons for the marks awarded. cao = correct answer only ecf = error carried forward bod = benefit of doubt

4 (a) Electrical energy (or work) transferred [to other forms] per unit [accept coulomb] of charge [passing between the two points] AO AO AO3 Total Maths Prac (b) Either: I in circuit =.4 60 (.0 -.4)() R Thermistor = x 0 = 640 [ ] () Or: R T = 9.6().4 60 () or () [= 5.0 ma] [ecf on I] x 60.4 () 60 R T 3 3 = 640 [ ] () (c) (i) [Resistance of thermistor decreases as temp increases] pd across thermistor decreases () So pd across fixed resistor increases because: Either - ratio of pds across potential divider changes Or - total pd must =.0 V (or equivalent) () [Resistance of thermistor decreases as temp increases] so circuit current increases () So pd across fixed resistor increases because V = IR and R is constant or V I ()

5 (ii) At 30 C, R thermistor = 480 from graph () 60 V cooling system = ( ).0 () [ecf on R Thermistor ] = 3.0 [V] () 60.8 =.0 () ( R 60) Thermistor R Thermistor = 56 [ ] () Corresponds to 5 C from graph () AO AO AO3 Total Maths Prac I.8 = [A] () R R = 56 [ ] () T Corresponds to 5 C from graph () I.8 R = [A] () 60 At 30 C, R thermistor = 480 from graph () I 9. T = 0.09 [A] () 480 At 30 C, R thermistor = 480 from graph () V I = [A] () R ( ) V = IR = = 3 [V] () Final mark for all methods - Valid conclusion consistent with answer: i.e. Claim incorrect - system activated at θ < 30 C ()

6 (d) More effective at 0 C 0 C (no mark) Because: Steeper gradient / larger change in resistance () Greater sensitivity in this range / greater [fractional] change in R per C change in temperature or over the same temperature range) () AO AO AO3 Total Maths Prac Question total

7 (a) (i) For Left Hand Combination: parallel (RHS seen in any correct form e.g. R R R R 4 R )() AO AO AO3 Total Maths Prac = R () R 3R Total R = + R or seen () Alternative solutions possible e.g. Sum of top and bottom branch = R () Then parallel branch = R () Total R = R + R () (b) (ii) Right hand resistor circled () Greatest current / greatest voltage () Correct substitution into l RA x 0 x 50 x 0 x 0.5 x 0 i.e. () -6.0 x 0 l = 0.0 [m] () (ecf on slip in powers of 0) 5

8 (c) (i) n- free electron density. Accept- number of free electrons per unit volume or per m 3 (or equivalent) (ii) Ratio Value Explanation n X Wires made of the same material ny I X Wires in series I Y v X 0.5 Correct explanation based on A xv x = A yv y v Y e.g (d) d v x = v y AO AO AO3 Total Maths Prac 3 (iii) Award mark for each correct row R l A substituted into P = I R i.e. xlx P I l A () I I zlz Px and Pz (or equivalent) - can award st mark Ax Az from one of these expressions lz A x = 4A z and l x and ρ x = ρ z to show: () P z P = 4 () x Question total

9 AO AO AO3 Total Maths Prac 3 (a) Electrons (or negative charges) are deposited on Z [and this plate becomes negatively charged] () Electrons (or negative charges) are removed from Y [and this plate becomes positively charged] () (b) (i) Initial pd across capacitor = pd of cell (by implication) and correct application to show R or I (ii) Reference to resolution of voltmeter () which is too small to be plotted () (on given scale) (iii) Error bars [are ± s] (iv) Appropriate (corresponding) values from graph e.g. V 0 = 6 V, V = 4 V, t = 3 s () t CR Correct algebra [ V V e ] to show t = 3 [s] () 0 Time constant = 0.37 V 0 stated or implied or V =.[] V () Time constant = 3 [s] () T ½ = 0.69RC () RC = 3 [s] () V 0 Initial gradient = 6 (tangent at t = 0 intercepts RC 33 time axis at t = 33 s) () RC = 6 x 33 = 33 [s] () 6 7

10 AO AO AO3 Total Maths Prac (v) Application of time constant = CR () 3 C = 47 F (ecf in t or candidate value used) () % uncertainty calculated as 3% + 3.% = 6.% () Absolute uncertainty = ± 30 F So: 470 ± 30 F (or 0.47 ± 0.03 mf) () consistency of sig figs (vi) t CR Correct substitution into V V e () 0 V shown =. [V] () Reference to continued graph line going through this point () Question 3 total

11 AO AO AO3 Total Maths Prac 4 (a) Horizontal line[s] with direction indicated from X to Y (b) (i) Substitution into F Ve -9 d shown: 800 x.6 x 0 () x 0 F = [N] () (ii) [Gain in E k = Work done by field] Gain in E k = () (ecf on F) Gain in E k = J unit mark () W = () W = J unit mark () [Accept 800 ev unit mark] (iii) x = ut + ½ at and u = 0 (all possible by implication) () a F and substitution step: ecf on F m (c) e.g. t x 3. x 0 x 9. x x 0-4 () t = [s] () ½ mv = to calculate v () ( ) Application of x u v t () ecf on v t = [s] () Ve F doubled (explained from d ) () W = F d so no change () 3 3 Accept: W = QV and Q stated to be constant () so W remains unchanged () Question 4 total

12 AO AO AO3 Total Maths Prac 5 (a) (i) Advantage Disadvantage Ben (ruler) Sarah (rod) Easy to use/convenient / quicker Diameter measured accurately / greater accuracy Inaccurate [only to ± mm] / reference to parallax errors / difficulty in supporting ruler / may touch spheres Diameter/radius of spheres need to be known beforehand / difficult to judge one complete rotation / difficult to measure angle [of rotation] / difficult to setup / thread overlapping 4 mark - one response required from each cell. (ii) Any () from: Pins/markers on ruler Marker on cylinder Measure diameter of spheres Mark point at centre of each sphere and use a travelling microscope to measure the separation Fix the ruler close to spheres Smaller diameter rod Use of Vernier calipers (for Ben) Don't accept repeat readings

13 (b) (i) F = N () 9 9 x 0 QQ Use of F () r Q Q = [C ] () (ii) Q = ( ) / determined (or use of ) = C () Area under graph calculated: = C () Area, Q = () So QQ = ( ) = C () (iii) x 0 n = electrons ecf on Q -9.6 x 0 AO AO AO3 Total Maths Prac Question 5 total 7 3 6

14 6 (a) Line drawn from Sun to planet... ()...will sweep out equal areas reference to A = A = A 3 ()...in equal time intervals / 6 months () (b) mv GMm = r r () r v T () Substitution and clear algebra step shown () Or: GMm mr () r () T Substitution and clear algebra step shown () AO AO AO3 Total Maths Prac (c) (i).45 years = [s] () v Substitution into c i.e = v x 0 () v = [m s - ] () vt x 0 x x 0 r = =.34 0 [m] () Alternative for see (ii) Assumption CoM at/near centre of neutron star or M much greater than M () Either: M vr () G 4 (3.09 x 0 )x.34 x 0 M (substitution) () ecf on v x 0 M = [kg] and valid conclusion () 4 4

15 (ii) 3 4 r M () GT AO AO AO3 Total Maths Prac 3 4π x(.34 x 0 ) M (substitution) () ecf on T x 0 x(4.573 x 0 ) M = [kg] and valid conclusion () Question 6 total

16 7 (a) (i) Potential at infinity = 0 () Work done on object to get to infinity, [therefore initial energy must be negative] (or equivalent) () (ii) I Zero () No change in potential (or on same equipotential ). Do not accept r unchanged unless reference made to potential unchanged () II V = (-.79) (-.3) () Change in gravitational E p = [-]44 M[J] () AO AO AO3 Total Maths Prac (iii) V at Moon surface calculated = -.8 M[J] per kg () Loss in E p = m 0.63 MJ (ignore ve sign) () ecf on V at Moon surface ½ mv = m () v = ( ) / [m s - ] =. 0 3 [m s - ] () E p at Moon = -846 M[J] Total energy at D = -657 M[J] () Loss in E p (gain in E k ) = 89 M[J] () v =. 0 3 [m s - ] () 4 4 4

17 (b) Benefits - Any () from: Easier for humans to survive on Moon if water present Help understand origin of Earth/Moon system To advance science Generate interest in science/space exploration Develop new technologies Create jobs Costs - Any () from: Funding could have been used to address Earth based issues Little current impact on society Pollution of Moon AO AO AO3 Total Maths Prac Question 7 total

18 8 (a) Plastic Deformation: P - Reference to dislocations or incomplete planes of atoms P - Applied forces break bonds near to dislocations P3 - Dislocations slip P4 - Original bonds permanently broken and do not reform or crystal does not return to original form when force removed Increasing Strength: S- Foreign atom S- Reduce grain size or increase number of grain boundaries S3- Further dislocations S4- Reason - how they work - inhibit dislocation movement AO AO AO3 Total Maths Prac marks Comprehensive description including both plastic deformation and increasing strength typically 6 or more points covered. There is a sustained line of reasoning which is coherent, relevant, substantiated and logically structured. 3-4 marks Comprehensive description of either plastic deformation or increasing strength or brief description of both plastic deformation and increasing strength typically 4-5 points covered. There is a line of reasoning which is partially coherent, largely relevant, supported by some evidence and with some structure. - marks Brief description of either plastic deformation or increasing strength - 3 points covered. There is a basic line of reasoning which is not coherent, largely irrelevant, supported by limited evidence and with very little structure. 0 marks No attempt made or no response worthy of credit. 6

19 (b) (i) CSA = ( ) = [ ] () Gradient from graph = or use of a point from the straight portion () AO AO AO3 Total Maths Prac Young Modulus = grad l A shown to be. 0 Pa () (ii) 0.% strain corresponds to an extension of 4.4 mm () Area under graph calculated = ½ () [W = 0.04 J ] (iii) Straight line from end of graph (between 6 8 mm) parallel to original line to x-axis. Tolerance: x-axis intercept between mm. Question 8 total

20 A LEVEL COMPONENT : ELECTRICITY AND THE UNIVERSE SUMMARY OF MARKS ALLOCATED TO ASSESSMENT OBJECTIVES Question AO AO AO3 TOTAL MARK MATHS PRAC TOTAL Eduqas GCE A Level Physics Component MS Summer 07 8

GCE AS MARKING SCHEME

GCE AS MARKING SCHEME SUMMER 206 PHYSICS AS - Unit 2420U0/0 INTRODUCTION This marking scheme was used by WJEC for the 206 examination. It was finalised after detailed discussion at examiners' conferences