COMPONENT 2 ELECTRICITY AND THE UNIVERSE MARK SCHEME GENERAL INSTRUCTIONS

Transcription

1 A LEVEL PHYSICS Specimen Assessment Materials 00 COMPONENT ELECTRICITY AND THE UNIVERSE MARK SCHEME GENERAL INSTRUCTIONS The mark scheme should be applied precisely and no departure made from it. Recording of marks Examiners must mark in red ink. One tick must equate to one mark (except for the extended response question). totals should be written in the box at the end of the 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.

2 A LEVEL PHYSICS Specimen Assessment Materials 0 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

3 A LEVEL PHYSICS Specimen Assessment Materials 0 (a) n number of free electrons per unit volume [accept m -3 ] () Accept free electron density v drift velocity [of free electrons] () (b) (i) mass = x 0-6 () [= kg] x0 5.5 = electrons () (ii) 3 5.x0 (ecf ) n () [= m -3 ] 6 4.0x0 (ecf ). (subst) () [ecf on n].8x0 x.0x0 x.6x0 v v = [m s - ] () [N.B. use of for n 7.3[5] [m s - ] 3 (c) Correct strategy determining n for each material () n calculated correctly for each material () A B C Therefore B made from different material () Alternative solution: Correct strategy realisation that e is constant () I calculated correctly for each material () Av Therefore B made from different material () 3 total

4 A LEVEL PHYSICS Specimen Assessment Materials 03 (a) (i) Diagram or set-up D Wire shown along with movable contact (accept crocodile clip). D Metre ruler. D3 Method for measuring resistance shown e.g. correct circuit with ohmmeter or correct circuit with voltmeter and ammeter appropriately connected. D4 Micrometer or digital callipers or vernier scales. Method M Vary length. M Measure resistance (or I and V). M3 Measure diameter or thickness. M4 Repeat readings of resistance and length and the mean values obtained. M5 Repeat readings of diameter (or thickness) and the mean value obtained. M6 Diameter (thickness) measured in different places along the wire. M7 Measure zero error of ohmmeter. Results R - Plot graph of resistance against length. R Measure gradient. R3 Intercept should be zero. R4 - RA quoted. l R5 - = gradient area R6 Area = r where r d 5-6 marks All of D D4 present. All of M M3 and M5 and either M6 or M7 present. All of R, R, R4, R5 and R6 present

5 A LEVEL PHYSICS Specimen Assessment Materials 04 There is a sustained line of reasoning which is coherent, relevant, substantiated and logically structured. 3-4 marks Expect any 3 from D D4. Expect M, M and M3. Expect R, R and R4. There is a line of reasoning which is partially coherent, largely relevant, supported by some evidence and with some structure. - marks Expect any from D D4. Expect M and M. Expect R. 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. (ii) Substitution into R l A () = [ m] () (b) Because R [for equivalent lengths] () A Therefore there is an increased resistance [per metre] () Therefore higher values of R for equivalent lengths lead to a steeper gradient () Resistivity does not change because the material of the wire does not change () 4 4 total 7 3 0

6 A LEVEL PHYSICS Specimen Assessment Materials 05 3 (a) (i) V is the terminal pd or clear explanation in energy terms: energy per coulomb transferred in the external circuit / [N.B. per coulomb / per unit charge required on one of (i) and (ii) if energy explanation given] (ii) pd across the internal resistance [accept lost volts bod ] / energy per coulomb lost / dissipated in the internal resistance / cell (b) (i).4 [V] (ii) 0.4 [Ω] [allow ecf from (b)(i)] (c) R Tot = (ecf) () I = [A] (ecf) () (d) At (0,0) there is no pd and no current () At (,6) as n becomes large becomes negligible () E Therefore equivalent to emf i.e. r () r Between (0,0) and (,6) becomes less and less significant so I increases () 4 3 total

7 A LEVEL PHYSICS Specimen Assessment Materials 06 4 (a) A 0 C used d 8.85x0 x x0 3 () Answer = 4. nf UNIT mark () (b) (i) 4.95 [μc] ecf (ii).97 [mj] ecf (c) 0 t Q Q e RC used or T RC ln used () Logs taken correctly e.g. T RC ln () Answer =.9 [ms] () ecf ln Q Q 0 t or substitute into RC 3 3 (d) Shorter since energy proportional to Q

8 A LEVEL PHYSICS Specimen Assessment Materials 07 4 (e) (i) V Correct circuit diagram e.g. as shown () Charging and discharging explained () e.g. switch left for charging, right for discharging Measure V at various (regular) time intervals () Use data logging because too quick for human () (ii) Plot a graph of ln V against t () Should be a straight line () Negative gradient / gradient = / positive intercept () RC total

9 A LEVEL PHYSICS Specimen Assessment Materials 08 5 (a) (b) r Correct use of sin i.e. = 0. 75sin0 () So r x0. 75sin0 = 0.6 [m] ( nd mark implies first) () F T sin0 () 3 also mg T cos0 where m kg () or equivalent Qq Use of F 4 r Q 0 4 0(0.6) F and comment () 4 4 (c) Attempt at approximation of uncertainties () Obtaining uncertainties for all 3 measurements () Identification of source of largest uncertainty (i.e. the angle) () Therefore a suitable improvement suggested () 4 4 (d) Oscillation about equilibrium point or mid-point () Air resistance acts on balloons () [Heavily] damped oscillations or [over] damped SHM or critical damping () 3 5 total

10 A LEVEL PHYSICS Specimen Assessment Materials 09 6 (a) (i) F GM N kg - m r Use of formula () Answer + UNIT () (ii) GM [J kg - ] r Use of formula () Answer () (b) Good strategy chosen involving the use of Kepler s law () 7 T ( 3. 6x0 ) 3 3 = Earth and r (. 5x0 ) 8 T ( 3. 74x0 ) 3 3 = Jupiter calculated () r ( 7. 79x0 ) 3 3 r (. 5x0 ) OR 7 = Earth and T ( 3. 6x0 ) 3 3 r ( 7. 79x0 ) 8 = Jupiter calculated T ( 3. 74x0 ) Conclusion that the agreement is good / excellent [this must be consistent with the calculations shown] () 3 6 total

11 A LEVEL PHYSICS Specimen Assessment Materials 0 7 (a) Starting from E FL 0 extension of bar of cross-sectional area A Ax L0 F shown to be and extension of bar of cross-sectional area A AE L0 F shown to be AE () Attempt at adding extensions made () Convincing algebra () e.g. common denominators shown: FL0 FL0 3FL0 4 4 AE AE 4AE (b) (i) Line drawn correctly i.e. gradient/steepness as given line (ii) Re-arrange for A bar, A bar or combination () Correct force-extension combination for each of above () Answer = 0 [N m - ] () (iii) Use of T m k () 400 Gradient = () 6 8x0 Period = 4 m[s] () total

12 A LEVEL PHYSICS Specimen Assessment Materials 8 (a) Absorbs all radiation (don t accept all wavelengths ) falling on it. Accept emits more radiation than any other body [at that temp] / perfect emitter / perfect absorber (b) (i) A = 4 ( ) or [m ] () P = area attempt () P = [W] with appropriate comment on consistency () [ for slips of n, 0 n ] Accept other alternatives e.g. finding P from A and T or finding A from P and T 3 3 (ii) 3.9x0 max or by implication () 5790 max = 500 n[m] No ecf () Graph goes through origin with zero slope and acceptable at high Peaks at approximately 500 nm ecf () 4 8 total

13 A LEVEL PHYSICS Specimen Assessment Materials 9 (a) (i) T = = [s] () TvS Use of rs or equivalent () = [m] () 3 3 (ii) Use of T GM 3 S d () 4 Assumption M S >> M P () = 3. 0 [m] () 3 3 (iii) r Use of M p Ms r d () = [kg] ()

14 A LEVEL PHYSICS Specimen Assessment Materials 3 9 (b) (i) Modern data has smaller random errors or points closer to the line of best fit () Modern data (or second graph) has no negative velocities (or blue shift) () Modern data is correct over a far larger range of distance ( Mpc 500 Mpc) () All of the above points can be made in reverse e.g. 99 data is very noisy or 99 data has data showing blue shift etc (ii) Hubble's constant is the gradient () Answer = 64 [± ] () 3 3 (iii) Age of the universe = H () 0 Answer = 5.3 billion [years] (ecf) () 9 total

15 A LEVEL PHYSICS Specimen Assessment Materials 4 COMPONENT : ELECTRICITY AND THE UNIVERSE SUMMARY OF MARKS ALLOCATED TO ASSESSMENT OBJECTIVES AO AO AO3 TOTAL MARK MATHS PRAC TOTAL