‘a’ Level Maths P4 Nov 2007 Marking Scheme

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2007 question paper 9709 MATHEMATICS 9709/04 Paper 4, maximum raw mark 50 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began.All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes.

CIE is publishing the mark schemes for the October/November 2007 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.Page 2 Mark Scheme GCE A/AS LEVEL – October/November 2007 Syllabus 9709 Paper 04 Mark Scheme Notes Marks are of the following three types: M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e. g.

by substituting the relevant quantities into the formula.Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). Mark for a correct result or statement independent of method marks.

A B • When a part of a question has two or more "method" steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated.The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. The symbol v implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously "correct" answers or results obtained from incorrect working.

Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2. • • The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e. g.

wrong working following a correct form of answer is ignored. • Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise.For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s. f.

, or which would be correct to 3 s. f. if rounded (1 d. p.

in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9. 8 or 9.

81 instead of 10. • © UCLES 2007 Page 3 Mark Scheme GCE A/AS LEVEL – October/November 2007 Syllabus 9709 Paper 04The following abbreviations may be used in a mark scheme or used on the scripts: AEF AG Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) Correct Answer Only (emphasising that no "follow through" from a previous error is allowed) Correct Working Only - often written by a ‘fortuitous' answer Ignore Subsequent Working Misread Premature Approximation (resulting in basically correct work that is insufficiently accurate) See Other Solution (the candidate makes a better attempt at the same question) Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) BOD CAO CWO ISW MR PA SOS SR Penalties MR -1 A penalty of MR -1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become "follow through v" marks.MR is not applied when the candidate misreads his own figures - this is regarded as an error in accuracy. An MR -2 penalty may be applied in particular cases if agreed at the coordination meeting. This is deducted from A or B marks in the case of premature approximation.

The PA -1 penalty is usually discussed at the meeting. PA -1 © UCLES 2007 Page 4 Mark Scheme GCE A/AS LEVEL – October/November 2007 M1 A1 M1 A1 M1 Syllabus 9709 Paper 04 1 DF – 550 = 900x0. 2 [P = 730x30 ? 1000] P = 21. 9 2 s1 = 12. 5t – ? gt2, s2 = ± ? gt2 or (12. 5 – gt)2 = 12.

52 – 2gs1 and (gt)2 = 2gs2 2 2 [12. 5t – ? gt + ? gt = 10] t = 0. 8s or 2s1 = 25 2 ? 0 . 2 s 1 ? ( 20 ? 2 s 1 ) (or better) Height is 6. 8mFor using Newton’s second law (3 terms) For using P = (DF)v 4 For applying s = ut + ? at2 or (u + at)2 = u2 +2as with a = ± g (either particle) A1 M1 A1 For using s1 + s2 = 10 A1ft 5 ft for 12. 5t – 5t2 or 10 – 5t2 with candidate’s t (requires both M marks) For stating F2 = 72 + 42 directly or for resolving in the i and j directions and eliminating ? or F Allow 8.

07 from 4 ? sin29. 7o For stating tan ? = 4/7 directly or for substituting for F or for ? into 7 = Fcos ? or 4 = Fsin ? Allow 29. 8 from sin-1(4 ? 8. 06) SR for candidates who mix sine and cosine (max 3/4) Fsin ? = 7, Fcos ? = 4 F2 = 72 + 42 M1 M1 For tan ? = 7/4 A1 For F = 7 and ? = 60. o Any equivalent form 3 [7 = Fcos ? and 4 = Fsin ? F2 = 72 + 42 (or tan ? = 4/7)] (i) F = 8.

06 [7 = 8. 06cos ? or 4 = 8. 06sin ? ] (or 7 = Fcos29. 7o or 4 = Fsin29. 7o) M1 A1 M1 ? = 29. 7 A1 4 (ii) Magnitude 7 N Direction opposite to that of the force of magnitude 7 N 4 (i) [? mv2- ? m72 = mgx5] Speed is 12.

2ms-1 B1 B1 2 M1 M1 A1 For using KE = ? mv2 For equation from KE gain = PE loss (3 terms) 3 SR for candidates who treat AB as straight and vertical (max 1mark out of 3) v2 = 72 +2g5 v = 12. 2 B1 For using WD = PE loss – KE gain or WD = KE at B in (i) – actual KE at B ft wrong v in part (i) or for 12. 2 scored by B1in (i) This mark is not available if v = 12. is used, having been scored by B1 in part (i) SR for candidates who treat AB as straight and vertical, and resistance as constant (max 1mark out of 3) a = 7. 2 ms-2, R = 0. 98 N, WD = 4.

9 J B1 SR for candidates who write ‘Resistance =’ instead of ‘WD =’ (max 2/3) 0. 35x10x5 – ? 0. 35(112 – 72) or ? 0. 35(12.

22 – 112) seen B1 Answer 4. 9J (NB J seen) B1 (ii) WD = 0. 35x10x5 – ? 0. 35(112 – 72) or WD = ? 0.

35(12. 22 – 112) Work done is 4. 9 J M1 A1ft A1 3 © UCLES 2007 Page 5 Mark Scheme GCE A/AS LEVEL – October/November 2007 M1 A1 M1 A1ft M1 A1ft A1 3 4 Syllabus 9709 Paper 04 5 (i) R = Tsin60o [F = W + Tcos60o ] F = 40 + Tcos 60o (ii) 40 + 0. 5T = 0.

7x0. 66T T = 377 For resolving horizontally (normal force must have a horizontal component) For resolving vertically (allow if normal force is not horizontal but equation must contain F , W and T) ft – allow F = 40 + Tsin 60o following R = Tcos60o For using F = µ R Any correct form ft unsimplified with candidate’s F(T) (with 2 terms) and R(T) 6 (i) [2. 2 = 1. 8 + 0. 004t] Time taken is 100s (or Distance is 200 m) Distance is 200 m (or Time taken is 100s) (ii) (a) s = k(100t2 – t3/3) (+C) [k(100x1002 – 1003/3) = 200] k = 0. 0003 (b) Speed is 3 ms-1 (iii) M1 A1 A1ft M1 A1 DM1 A1 B1ft M1 M1 4 1 3 For using v = u + at (or v2 = u2 + 2as) ft s = 2t or 1.

8t + 0. 02t2 (or t = s/2) For integrating v(t) to find s(t) For using s(0) = 0 (may be implied) and s(100) = 200 ft candidate’s t and/or k. For straight line segment, v(t) +ve and increasing throughout (including at t = 0) For parabolic segment through origin, with +ve slope Parabolic segment has decreasing slope; sketches correct relative to each other (line crosses curve once) A1 3 Depends on both M marks © UCLES 2007 Page 6 Mark Scheme GCE A/AS LEVEL – October/November 2007 R T Syllabus 9709 Paper 04 7 (i) B1 1 The components F and R may be represented by a single contact force, which must be shown at an acute angle to the downward slope. F W (ii) T – F – 0. 13g (16/65) [T – F – 0. 13g (16/65) ; 0] T – F ; 0.

32 (iii) R = 0. 3g(63/65) or 0. 13g cos14. 25… (= 1. 26) F = 0. 6 x 1.

26 (= 0. 756) M1 A1 M1 A1 B1ft M1 M1 4 For finding the resultant upward force (RUF) (3 terms required) For use of RUF ; 0 (since P starts to move upwards). AG ft 0. 13g cos 75. 7….

. For using F = µ R For applying Newton’s second law to P (4 terms required) or to Q (3 terms required) or for using WQ - WPsin ? - F = (mP + mQ)a ft1. 26 instead of 0. 32 following a consistent sin/cos mix throughout (i) and (ii) For substituting for F and solving for a. 6 T – F – 0.

32 = 0. 13a and 0. 11g – T = 0. 11a or 0. 11g – F – 0. 32 = (0.

13 + 0. 11)a Acceleration is 0. 1 ms-2 A1ft M1 A1 © UCLES 2007