Mark Scheme (Final Standardisation) Summer 2007

Transcription

1 Mark Schee (Final Standardisation) Suer 007 GCE GCE Physics (675/0) Edexcel Liited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

2 675 Unit Test PHY 5 final version, agreed and ipleented at standardisation.. a) Show that πr π See v OR ω T T Substitution of ( )s or 86400s for T (giving 7.7-5, no u.e.) Unit of ω s - / rad s - b) Height above Earth s surface Stateent / use of E v OR E rω r r r [Equations ay be given in ters of accelerations rather than forces] [Third ark (fro below) ay also be awarded here if (r E +h) is used for r] Correct value for r, i.e. 4.() 7 Use of h their r - R E Correct answer (.58.60) 7 [no ecf] E v r r E v r r ω r r E 4. 7 E ( ωr) r ω ω r 6.67 N kg ( s ) 4 kg 4 h Total 7 arks

3 675 Unit Test PHY5. Add to diagra. Arrows at A and B, both pointing directly away fro the nucleus. [Arrow end (head or tail) need not touch A /B, but direction ust be correct. Gauge by eye, accept dotted construction lines as indication of intent] Calculation of force QQ kqq Use of F or F 4πε 0r r [ignore error/oission of and/or 79 or e or.6-9 for this first ark, providing nuerator clearly has a product of charges and denoinator a distance value squared. Ignore power of errors in values of Q or r].6-9 C and C seen (consequential ark, dependent upon correct use of equation previously) Correct answer.6.7 N F Q Q 4πε o r (79.6 4π 8.85 C) (.6 F ( C) ).6 N Effect on otion of α Slows down [decelerates] and then speeds up again [accelerates]. (both needed) [accept slows down at A and speeds up at B] Total 5 arks

4 . a) How capacitors are connected 675 Unit Test PHY 5 Box A in parallel [Accept diagras] Box B in series [N.B. If A in series and B in parallel, ax in explanation section below] Explanation Answer with no word or sybol reference to energy scores /4 ax. Answer with no reference to any relevant forulae scores /4 ax. More energy stored in A [ A ay be iplied by arguent] The sae p.d. ( V ) (So) C A > C B [or deonstration by nuerical exaple] Use of* W ½CV [e.g. ½C A V > ½C B V ] OR use/stateent of E ½QV AND QCV Use of* either equivalent capacitance forula [correctly stated; ay be word equation] * i.e. Referred to as part of explanation. Do not credit bald transcription of equations given in the list at the back of the paper without context, nor as arginalia]. [Award arks for correct, non-contradictory stateents even if the candidate has given the wrong cobinations at a(i), up to a axiu of arks] Max 4 b) Addition of large resistor in discharging circuit Valid observation in ters of brightness or duration of illuination Supporting explanation in ters of circuit behaviour [Max ark if explanation does not support observation, or is internally contradictory, or if description does not include a visual observation] c) Addition of large resistor in charging circuit Valid observation in ters of brightness or duration of illuination Supporting explanation in ters of circuit behaviour [Max ark if explanation does not support observation, or is internally contradictory, or if description does not include a visual observation] Total 9 arks

5 675 Unit Test PHY5 4. Explanation of kicking Answer with no reference to oents scores 4/5 ax. Ignore references to electroagnetic induction/lenz s Law QoWC (Fleing s) left hand rule / agnetic fields interact/cobine/overlap [not repel, nor interfere ] / reference to current flowing in agnetic field /catapult field Force acting on the wire linked to oent [not just pivoting ] (about P) Force to right / anticlockwise oent [detail of direction] When wire leaves ercury, current 0 / force 0 / oent 0. [not just circuit is incoplete ] Idea that wire s weight produces a oent (returning it to ercury) Show that Use of oent equation, i.e N F d [accept any nuerical value for d between.5 (c) and.5 (c)] Use of 6 - for d Max 5 Correct answer 8. - (N) [no u.e.] [Reverse arguent scores /] 4 Moent.0 N F 8. d ( ) Circuit current 5 N Use of F BIl [or correct rearrangeent] with l 9c [Ignore powers of. No ecf for their force if different; beware use of ] Answer./. A F F BIl I Bl 8. N I 4.0 T 9.0.A [8x - N.A] Total arks

6 675 Unit Test PHY5 5. a) Direction of e..f.? Hub - and Ri +. [Allow ark for either on its own, but not if contradicted.] Why a constant e..f.? Reference to flux cutting / rate of change of flux / change of flux linkage due to spoke otion / spokes oving at right angles to field / Reference to Faraday s Law Constant rate of spin iplies constant rate of flux cutting. [Link ade clear] [continuous process does not ean constant rate] The tie for one revolution BA Use of ε with A recognisable as area of a circle t [ignore power of errors for e..f. and radius values, and inclusion of N4] Correct substitution of all values [ but only N acceptable here] Correct answer s [t 7.6s scores /; t.s scores 0/, t 0.64s scores / here] ϕ ε t BA t.8 t t 5 BA ε T π 5 ( 0 ) 0.7 s 6 V Alternative answer Use of ε Blv with v (ean) velocity of spoke. v.98 s - Hence ri velocity s -. πr π 0. t 0.6s s v RIM [t 0.6s scores / here] () () ()

7 What effect? (i) Reduced [accept halved ] AND Rate of flux cutting is reduced / Fewer field lines are being cut / Coponent of Earth s field perpendicular to the wheel is less /Flux through wheel is less / Area of wheel perpendicular to field is less / Wheel is no longer perpendicular to the field [do not credit answers suggesting changes in the field strength itself] (ii) Increased / increasing AND Rate of flux cutting [etc.] would be increasing (iii) (Reduced to) zero [but not very sall / negligible, etc.] AND No flux cut by spoke(s) / No coponent of the Earth s field perpendicular to the wheel / No flux through wheel / Wheel is spinning parallel to the field / in plane of field [but not just Ф 0, nor otion is not perpendicular to field ] [Allow / for three correct stateents of ε outcoe without any explanation, but only if score would otherwise be zero.] [Disallow breaking for cutting on first occasion in entire question, but allow, ecf, thereafter] Total 9 arks