9231 FURTHER MATHEMATICS

Transcription

1 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the May/June 5 series 9 FURTHER MATHEMATICS 9/ Paper (Paper ), maimum raw mark This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the eamination. It shows the basis on which Eaminers were instructed to award marks. It does not indicate the details of the discussions that took place at an Eaminers meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Eaminer Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 5 series for most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademark of Cambridge International Eaminations.

2 Page Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 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. A B 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. 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 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: B or A means that the candidate can earn or. B// means that the candidate can earn anything from to. 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 s.f., or which would be correct to s.f. if rounded ( 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 9.8 instead of. Cambridge International Eaminations 5

3 Page Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 The following abbreviations may be used in a mark scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so etra 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 Approimation (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) Penalties MR A penalty of MR 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 " marks. MR is not applied when the candidate misreads his own figures this is regarded as an error in accuracy. An MR penalty may be applied in particular cases if agreed at the coordination meeting. PA This is deducted from A or B marks in the case of premature approimation. The PA penalty is usually discussed at the meeting. Cambridge International Eaminations 5

4 Page 4 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Qn & Part Solution Marks (Award M for subtracting 9 or here.) 4 7k 8 k 4 k Add st and rd y ( Same as nd ) (OE) Set t (for eample) (OE) 7 5 y t, z t (OE) many forms 4 4 a > 5 (given) H is true. Assume H k is true for some positive integer k, i.e. a k 5 δ, where δ >. 4ak 5 4ak 55ak (4ak 5)( ak 5) a k 5 5 >, 5ak 5ak 5a a k > 5 k Or δ δ a k (5 δ ), 4 δ (...) for < δ < δ δ ( δ ) [ a k > 5, ( δ [ 5 is trivial). 5 H k H k and H is true, hence by mathematical induction, the result is true for all n Z (N.B. The minimum requirement is true for all positive integers.) 5 ak ak ak a 5 5 < and ak > 5 a k k k ak < ak a < a k MA MA (4) Total: 4 MA () M M AA (4) Total: 6 B B MA (M) (A) A (5) M A () Total: 7 4 (i) α β γ 7, αβ βγ γα, αβγ (Stated or implied by working.) ( αβ )( βγ )( γα) ( αβγ ) 9 B B (ii) (iii) α β γ 7 αβ βγ γα αβγ α βγ αβ γ αβγ αβγ α β γ αβ βγ γα αβγ αβγ MA MA (6) MA () Total: 8 Cambridge International Eaminations 5

5 Page 5 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Qn & Part Solution Marks 5 sin θ θ π ; Intersection at, π (Accept.5 for π.) MA () 6 7 C : Circle centre at pole and radius / C : Curve, appro. correct orientation, from (,) to (,π). Completely correct correct shape. π π sin θdθ 6 π π cos θ π dy dy 6 y 5y d d At (,) 6 8y ' 6 5y' y' (AG) 6( y" y' y') 5[( y') y y"] At (,) 8y " 5y" y" 6 maimum. (All previous marks required for final mark, i.e. CSO) n π π π n sin d [ n cos ] n cos d π π n n n n π n( n ) I n (LNR) I n n sin n( n ) sin d (LR) π π n [ ] I sin d cos I π, using reduction formula. 4 (AG) π I 4 ( π ) π π 4 (If I found by integration MA (if correct) then MA for I 4 from reduction formula.) B B B () MA MA (4) Total: 9 MA A A (4) BB MA A (5) Total: 9 MA MA A (5) B MA A (4) Total: 9 Cambridge International Eaminations 5

6 Page 6 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 n n r z( [ z ] ) 8 z z r z z z n z n (cos nθ i sin nθ ) cos nθ isin nθ z z cosθ i sin θ (cos θ i sinθ ) i sin θ Equating imaginary parts: n sin nθ sin nθ sin (r ) θ (AG) r sinθ sinθ MA MA A MA (7) Differentiating: n sin (r ) cos (r ) θ n sin nθ cos nθ cosec θ sin nθ cosec θ cotθ r π Putting θ : n n π π π π π π π (r) cos (r ) nsin cos cosec sin cosec cot r n n n n n (r ) π π π (r )cos cosec cot. (AG) r n n n MA dm A (4) 9 & & 4 t & y& 8t y 4 t s t t t 7 4 ( 8 ) d ( 8 ) 4.6 (CAO) 7 d y d y dt t( 4 t)( 4 t) dt 6t 4t 6tdt dt MV 5 8 8t t t ( 5.) 5 a y d b 5..6 (CAO) b a 4 4 Total: B B MA MA (6) M MA MA (5) Total: Cambridge International Eaminations 5

7 Page 7 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 eigenvalue is. MA () λ 4 : i j k 7 7 ~ (Or by equations) MA λ 6 : i j k 4 ~ 6 A () P D 4 6 B QAQ QPAP Q ; QPA (QP) QP (CAO) Eigenvalues are, 4 and 6 (same as those for A). (CAO) B B () M;A B B Eigenvectors are: 5 7, 5, 7. B (5) Total: dv dy (e) d y d d v d y dy d v d y dy y y y y d d d d d d y y y v v ~ v d d d 5 d d y d y d y d y d d (AG) m m 5 m ± i CF : e ( A cos B sin ) PI : v p q r v' p q v" p Equate coefficients: 5 p 5 4 p 5q 6 p q 5r 7 p, q, r GS : v e ( Acos Bsin ) B MA A (4) MA M M A A When y v and d y d v 4 d d Cambridge International Eaminations 5

8 Page 8 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Cambridge International Eaminations 5 A A ) sin cos ( ) cos sin ( ' B A e B A e v 4 B B B cos sin e y B M A A () Total: 4 (i) (ii) (a) (b) p, 8 λ λ q µ µ λ λ µ µ QP µ λ µ µ λ λ µ λ µ µ λ λ Solving : λ, µ 5 Whence: p 4, q , 6 AQ AP i j k (CAO) ; Area 5 (.5) AB Volume 5 (.5) uuur MA MA A MA A (8) B MA A MA (6) Total: 4

9 Page 9 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Alternative: for marks,4,5,6 and 7 in part (i) i j k 6 6 k λ k λ k µ 7 Solving; k, λ and µ 5 (If k missing, or assumed to be, deduct mark.) (MA) (A) (MA) Cambridge International Eaminations 5

10 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the May/June 5 series 9 FURTHER MATHEMATICS 9/ Paper (Paper ), maimum raw mark This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the eamination. It shows the basis on which Eaminers were instructed to award marks. It does not indicate the details of the discussions that took place at an Eaminers meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Eaminer Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 5 series for most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademark of Cambridge International Eaminations.

11 Page Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 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. A B 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. 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 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: B or A means that the candidate can earn or. B// means that the candidate can earn anything from to. 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 s.f., or which would be correct to s.f. if rounded ( 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 9.8 instead of. Cambridge International Eaminations 5

12 Page Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 The following abbreviations may be used in a mark scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so etra 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 Approimation (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) Penalties MR A penalty of MR 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 " marks. MR is not applied when the candidate misreads his own figures this is regarded as an error in accuracy. An MR penalty may be applied in particular cases if agreed at the coordination meeting. PA This is deducted from A or B marks in the case of premature approimation. The PA penalty is usually discussed at the meeting. Cambridge International Eaminations 5

13 Page 4 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Qn & Part Solution Marks (Award M for subtracting 9 or here.) 4 7k 8 k 4 k Add st and rd y ( Same as nd ) (OE) Set t (for eample) (OE) 7 5 y t, z t (OE) many forms 4 4 a > 5 (given) H is true. Assume H k is true for some positive integer k, i.e. a k 5 δ, where δ >. 4ak 5 4ak 55ak (4ak 5)( ak 5) a k 5 5 >, 5ak 5ak 5a a k > 5 k Or δ δ a k (5 δ ), 4 δ (...) for < δ < δ δ ( δ ) [ a k > 5, ( δ [ 5 is trivial). 5 H k H k and H is true, hence by mathematical induction, the result is true for all n Z (N.B. The minimum requirement is true for all positive integers.) 5 ak ak ak a 5 5 < and ak > 5 a k k k ak < ak a < a k MA MA (4) Total: 4 MA () M M AA (4) Total: 6 B B MA (M) (A) A (5) M A () Total: 7 4 (i) α β γ 7, αβ βγ γα, αβγ (Stated or implied by working.) ( αβ )( βγ )( γα) ( αβγ ) 9 B B (ii) (iii) α β γ 7 αβ βγ γα αβγ α βγ αβ γ αβγ αβγ α β γ αβ βγ γα αβγ αβγ MA MA (6) MA () Total: 8 Cambridge International Eaminations 5

14 Page 5 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Qn & Part Solution Marks 5 sin θ θ π ; Intersection at, π (Accept.5 for π.) MA () 6 7 C : Circle centre at pole and radius / C : Curve, appro. correct orientation, from (,) to (,π). Completely correct correct shape. π π sin θdθ 6 π π cos θ π dy dy 6 y 5y d d At (,) 6 8y ' 6 5y' y' (AG) 6( y" y' y') 5[( y') y y"] At (,) 8y " 5y" y" 6 maimum. (All previous marks required for final mark, i.e. CSO) n π π π n sin d [ n cos ] n cos d π π n n n n π n( n ) I n (LNR) I n n sin n( n ) sin d (LR) π π n [ ] I sin d cos I π, using reduction formula. 4 (AG) π I 4 ( π ) π π 4 (If I found by integration MA (if correct) then MA for I 4 from reduction formula.) B B B () MA MA (4) Total: 9 MA A A (4) BB MA A (5) Total: 9 MA MA A (5) B MA A (4) Total: 9 Cambridge International Eaminations 5

15 Page 6 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 n n r z( [ z ] ) 8 z z r z z z n z n (cos nθ i sin nθ ) cos nθ isin nθ z z cosθ i sin θ (cos θ i sinθ ) i sin θ Equating imaginary parts: n sin nθ sin nθ sin (r ) θ (AG) r sinθ sinθ MA MA A MA (7) Differentiating: n sin (r ) cos (r ) θ n sin nθ cos nθ cosec θ sin nθ cosec θ cotθ r π Putting θ : n n π π π π π π π (r) cos (r ) nsin cos cosec sin cosec cot r n n n n n (r ) π π π (r )cos cosec cot. (AG) r n n n MA dm A (4) 9 & & 4 t & y& 8t y 4 t s t t t 7 4 ( 8 ) d ( 8 ) 4.6 (CAO) 7 d y d y dt t( 4 t)( 4 t) dt 6t 4t 6tdt dt MV 5 8 8t t t ( 5.) 5 a y d b 5..6 (CAO) b a 4 4 Total: B B MA MA (6) M MA MA (5) Total: Cambridge International Eaminations 5

16 Page 7 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 eigenvalue is. MA () λ 4 : i j k 7 7 ~ (Or by equations) MA λ 6 : i j k 4 ~ 6 A () P D 4 6 B QAQ QPAP Q ; QPA (QP) QP (CAO) Eigenvalues are, 4 and 6 (same as those for A). (CAO) B B () M;A B B Eigenvectors are: 5 7, 5, 7. B (5) Total: dv dy (e) d y d d v d y dy d v d y dy y y y y d d d d d d y y y v v ~ v d d d 5 d d y d y d y d y d d (AG) m m 5 m ± i CF : e ( A cos B sin ) PI : v p q r v' p q v" p Equate coefficients: 5 p 5 4 p 5q 6 p q 5r 7 p, q, r GS : v e ( Acos Bsin ) B MA A (4) MA M M A A When y v and d y d v 4 d d Cambridge International Eaminations 5

17 Page 8 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Cambridge International Eaminations 5 A A ) sin cos ( ) cos sin ( ' B A e B A e v 4 B B B cos sin e y B M A A () Total: 4 (i) (ii) (a) (b) p, 8 λ λ q µ µ λ λ µ µ QP µ λ µ µ λ λ µ λ µ µ λ λ Solving : λ, µ 5 Whence: p 4, q , 6 AQ AP i j k (CAO) ; Area 5 (.5) AB Volume 5 (.5) uuur MA MA A MA A (8) B MA A MA (6) Total: 4

18 Page 9 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Alternative: for marks,4,5,6 and 7 in part (i) i j k 6 6 k λ k λ k µ 7 Solving; k, λ and µ 5 (If k missing, or assumed to be, deduct mark.) (MA) (A) (MA) Cambridge International Eaminations 5

19 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the May/June 5 series 9 FURTHER MATHEMATICS 9/ Paper, maimum raw mark This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the eamination. It shows the basis on which Eaminers were instructed to award marks. It does not indicate the details of the discussions that took place at an Eaminers meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Eaminer Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 5 series for most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademark of Cambridge International Eaminations.

20 Page Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Mark Scheme Notes Marks are of the following three types: M A B 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. 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 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: B or A means that the candidate can earn or. B// means that the candidate can earn anything from to. 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 s.f., or which would be correct to s.f. if rounded ( 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 9.8 instead of. Cambridge International Eaminations 5

21 Page Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 The following abbreviations may be used in a mark scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so etra 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 Approimation (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) Penalties MR PA A penalty of MR 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 " marks. MR is not applied when the candidate misreads his own figures this is regarded as an error in accuracy. An MR 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 approimation. The PA penalty is usually discussed at the meeting. Cambridge International Eaminations 5

22 Page 4 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Qn & Part Solution Marks i α β ii α 4αβ β p iii αβ αβ q iv α β r All four correct (Any two correct) Use of β α in ii, iii or iv 4 p α and α r p 4r (AG) and q (CAO) B (B) M MA B (6) Total 6 α α 8θ 8θ 4θ e 6e dθ 7e d α s θ (AEF) (LNR) 7 4θ 7 4α 7 e e α e K α.89 CAO MA dma dma (6) Total 6 k k H k : is true for some integer k. r ( r) k H is true. k k k k k (k ) k (k )(k ) (k )(k ) (k )( k ) k (k )(k ) [ k ] H k Hk (By Principle of Mathematical Induction) H n is true for all positive integers n. (This mark requires all previous marks.) r r ( ) B B MA A A (6) B () Total 7 Cambridge International Eaminations 5

23 Page 5 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Qn & Part Solution Marks 4 tan ( ) ( ) A ( ) ; tan B ( ) tan( A B) ( )( ) tan( A B) A B tan (AG) LHS (tan tan - tan [ n ] tan ) (tan (tan n tan tan ) K (tan [ n ] tan tan tan π π 5 In In dθ [ n ]). n tan [ n ] tan [ n ]) n sin( nθ) sin(n) θ sin(n ) θcosθ dθ cosθ cosθ (LR) π cos(n) θ sin(n ) θ dθ (n ) (LR) [] (AG) (n ) n π π π sin θ I [ ] dθ sin θ dθ cos θ [] [ ] cosθ 4 I I I π π 4 MA A () MA AA (4) Total 7 MA dma A (5) B MA A (4) Total 9 6 n n n n n ( z ) z z z L n Re {( n n n z n ) } cosθ cosθ cos nθ n L n ([ cos θ] isin θ) Re cos θ i sin θ cos θ n ( ) n ( ) n n n cosnθ cos ( θ) cos ( nθ ) Re [ ] [ ] [ ] n n cos ( θ) Re cos[ θ] i sin[ θ] n n cos cos θ θ L (AG) n n n n n Im{( z) } sin θ sin θ L sin nθ n n n sinθ sin θ L n n n sin nθ cos ( ) n θ sin ( nθ ) B MA MA A A (7) M A () Total 9 Cambridge International Eaminations 5

24 Page 6 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Qn & Part Solution Marks dy dy 7 y 8y ( st B for st and rd terms, nd B for rest including ). d d dy dy ( 4y) ( y) y when d d Substituting for y 4 5 ± Coordinates are (, ) and (, ) d d d d ( 4 y) 4 d d d d y y y y (OE) d y Substituting either (, ) or (, ) and d At (, ) y () ma. At (, ) y () min. λ 8 λ. λ λ λ λ λ P is (,, 4) (Accept position vector.) n i j k 5 w 4 BB B () M A MA dm A A (7) Total M A A () B MA () i Direction of PQ is 5 j 4 k 5 ~ 4 7 (OE) MA 5 Equation of PQ is r µ 4 7 [N.B. For methods not using w at most three B marks.] dma (4) Total Cambridge International Eaminations 5

25 Page 7 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Qn & Part Solution Marks 9 ( m 5)( m ) CF : Ae 5t Be PI : p sin t q cos t & p cos t q sin t & p sin t q cos t p sin t q cos t p cos t q sin t p sin t q cos t sin t cos t p q and p q p., q. t MA M A dma 5t t GS: e e. cos. sin A B t t CAO Initial conditions A B.. A B 5 and from 5 e t t & A Be. sin t. cos t 5 A B..9 5A B 5, e t t A B e. cos t. sin t CAO A B M A A () Total ( ) ( ) ( ) ( ) 9 ( ) ( ) ( ) ( ) ( ) > y > ( ) ( ) 9 > y < < y < ( ) 9 (Not hence gets B only.) 9 Turning points occur at equality, so, and, MA A () B B () BB () Asymptote is y 4. B () Intersection with aes at (, ) and Correct graph., 4, Intersection with asymptote at 4, 4 BB B () Total Cambridge International Eaminations 5

26 Page 8 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Cambridge International Eaminations 5 Qn & Part Solution Marks E (i) L Dim (V) MA A () (ii) 4 c b a i c b a ii c b a iii c b a i iii b a c and a c c and a. So vectors are linearly independent. Since, 4, has three linearly independent vectors, it forms a basis for V. M MA A B (5) (iii) W is not a vector space as it does not contain the zero vector. B ()

27 Page 9 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Qn & Part Solution Marks (iv) 4 y L z t 5 y 5y z y z t MA Alternatively: y α β γ (i.e. in V) z t 4 α β γ y α β 5γ y α β γ z α β γ z t α β 5γ t α 4β γ ( y) ( z t) Vector belongs to V iff y z t. (OE) So belongs to W iff y z t. (AG) (MA) A M A (5) Total 4 O 4 4 α 9 6 8α 8 α MA 6 7 A () λ 4 (i) 4 λ 6 6 λ M λ 6λ 6 ( λ 9)( λ )( λ 6) λ 6 and λ. MA MA (5) Cambridge International Eaminations 5

28 Page Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Qn & Part (ii) Solution λ 9 e (Award MA for any one λ 6 e and λ e and A for the other two). Marks MA A () M M( ae be ) ame 6ae bme aλ e bλ e be B B B () Total 4 Cambridge International Eaminations 5

29 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the May/June 5 series 9 FURTHER MATHEMATICS 9/ Paper, maimum raw mark This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the eamination. It shows the basis on which Eaminers were instructed to award marks. It does not indicate the details of the discussions that took place at an Eaminers meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Eaminer Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 5 series for most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademark of Cambridge International Eaminations.

30 Page Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Mark Scheme Notes Marks are of the following three types: M A B 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. 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 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: B or A means that the candidate can earn or. B// means that the candidate can earn anything from to. 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 s.f., or which would be correct to s.f. if rounded ( 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 9.8 instead of. Cambridge International Eaminations 5

31 Page Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 The following abbreviations may be used in a mark scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so etra 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 Approimation (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) Penalties MR PA A penalty of MR 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 " marks. MR is not applied when the candidate misreads his own figures this is regarded as an error in accuracy. An MR 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 approimation. The PA penalty is usually discussed at the meeting. Cambridge International Eaminations 5

32 Page 4 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Question Number Mark Scheme Details Part Mark Total Find T by equating dv/dt at t T to 6: 4T 4 6, T.5 M A Find radial component v /r of acceln. at t T: v /r (T 4T ) /.5 (M if T not given a value) (/) 4 [m s ] M A SR: Ma M (/4) if linear and angular confused 4 4 (i) Find ω from SHM eqn. d /dt ω at C:.65 ω, ω.65 or /6 B Find period T [s] from T π/ω: (ft on ω ) T π/¼ 8π (not 5.) B (ii) Find amplitude a [m] from v C ω (a ): 6 ω (a ) a 6 6, a M A Find time from C to M, e.g.: ω sin (/a) ω sin ½ or ω cos ( /a) ω cos ½ or ½T ω cos (/a) ω cos ½ M (AEF throughout) ω {.948 π/6 [.56]} or ω {.9656 π/ [.47]} or ω {π.76 π/ [.47]} A or or or 4.984;.67 [s] A; A Find v from conservation of energy: ½mv ½mu mga( cos θ ) M A Find R by using F ma radially: R mg cos θ mv /a B Eliminate v to find R: AG R mg( cos θ ) mu /a M A Find u or v in terms of cos θ when R : u ag( cos θ ) or v ag cos θ B EITHER: Replace cos θ in energy eqn with v u: 4u u ag ⅔(u ag) or u ag 8u M A OR: Find cos θ and substitute in energy eqn: [v /ag ] 4( cos θ ) cos θ cos θ 8/ 4u u ag ( 8/) (M A) Hence find u: u (ag/) or.46 (ag) A (i) Take moments for rod about B: W a cos W a cos (or with cos /) T a cos M A Hence find tension T: T 7W/ A (Can earn M A A if e.g. sin wrongly used) Find modulus λ using Hooke s Law: T λ (a a/5) / (a/5) λ (/7) (7W/) W/ M A 5 Cambridge International Eaminations 5

33 Page 5 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Question Number Mark Scheme Details Part Mark Total (ii) EITHER: Find horizontal component of force F at B: X T cos (7 /4) W or. W B Find vertical component: (X, Y ft on T) Y 4W T sin 9W/4 B Find magnitude of F: F (X Y ) ( 57/) W or.77[5] W B Find direction of F (AEF): Upward force at angle to AB of tan Y/X tan /7 (A if direction unclear) 6.6 or.69 radians M A OR: Find component along CB: F (4W T) sin 5W/4 (B ) Find normal component: (F, F ft on T) F (4W T) cos ( /4)W (B ) Find magnitude of F: F ( 57/) W or 77[5] W (B) Find direction of F (AEF): Upward force at angle to CB of tan F /F tan /5 (A if direction unclear) 6.6 or.5 radians (M A) OR: Find component parallel to string CA: ±F T 4W sin W/ (B ) Find normal component: (F, F ft on T) ±F 4W cos W (B ) Find magnitude of F: F ( 57/) W or.77[5] W (B) Find direction of F (AEF): Upward force at angle to AC of tan F /F tan 4/ (A if direction unclear) 66.6 or.6 radians (M A) 5 5 For A & B use conservation of momentum, e.g.: mv A mv B mu M (m may be omitted here and below) Use Newton s law of restitution (consistent signs): v B v A eu M Combine to find v B : v B ( e) u/5 A For B & C use conservation of momentum, e.g.: mv B mv C mv B M Use Newton s law of restitution (consistent signs): v C v B e v B M Combine to find v C and v B : v C ( e ) v B / ( e) ( e ) u/5 AG A v B ( e ) v B / ( e) ( e ) u/5 A Find ratios or values of v A, v B, v C from momentum: v A v B v C [ u] B Find e from first collision eqns, e.g.: v A ( e) u/5 u/ (or find e and then use v A v B ) or v B ½ (u u) or (⅓ e) u ( e) u/5, e ⅔ M A Find e from second collision eqns, e.g.: v B v C so ( e ) ( e ) or v C ( ⅔) ( e ) u/5 u or v B ( ⅔) ( e ) u/5 u/ e ½ M A Equate pooled estimate of σ to : ( 5 /N 6 /) / (N ) M A Formulate and solve relevant quadratic eqn. for N: N 65 N 5, N 5 M A 4 4 Cambridge International Eaminations 5

34 Page 6 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Question Number Mark Scheme Details Part Mark Total 7 Find Σ via sample mean : Σ 8 8 ½ (.7.) M A Find estimate of population variance s : t (s /8) ½ (..7) [.4] M Use of correct tabular value ( 96 leads to ): t 7,.975.6[5] A (to d.p.) s.645 or / or.54 A Find Σ from s : s {Σ (Σ ) /8}/7 (M for s { }/8) Σ /8. M A (a) (i) Find correlation coefficient r from r b b : r (.8.96).64 M *A (ii) State both hypotheses (B for r ): H : ρ, H : ρ > B State or use correct tabular one-tail r-value: r, 5%.549 *B State or imply valid method for reaching conclusion: Reject H if r > tab. value (AEF) M Correct conclusion (AEF, dep *A, *B): There is positive correlation A 4 (b) State or use relevant tabular two-tail r-value: r 6, 5%.497 (or r 5, 5%.54) M Find least possible value of n: n min 6 A SR M A for stating 6 without eplanation B for stating 5 without eplanation B for finding or stating one-tail result 8 9 (i) Relate P(X > ) to number of flaws (AEF): P(X > ) P(zero flaws in m) B Relate this to Poisson distn. (AEF): P (.8) e 8 A.G. B (ii) Find P(number of flaws ): P (.8 4) e (M if omitted) M A (iii) (a) Find or state distribution function F(): F() P(X ) P(X > ) e 8 B (b) Find or state probability density function f(): f() df/d 8 e 8 M A S.R. Deduct A if (a), (b) interchanged (c) Formulate equation for either quartile value Q : F(Q) e 8Q ¼ or ¾ M Find lower quartile Q : (AEF) Q. ln 4/ [.6 ] A Find upper quartile Q : (AEF) Q. ln 4 [.7] A Find interquartile range (allow Q Q ): Q Q [. ln ].7 A 4 Cambridge International Eaminations 5

35 Page 7 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Question Number Mark Scheme Details Part Mark Total Calculate gradient b in y y b ( ) : S y / 4.8 S / 6.96 (PA so ma 4/5 for.9, giving y 7.48) b S y / S.98 M A Find y when 7 from regression line of y on : y 56.9/ b (7 5.8/) M A (7 5.8) [ ] y 7.47 (allow 7.48 or 7.5) A SR If regression line of on y used: S yy /.89 b S y / S yy.7 (M) 7 5.8/ b (y 56.9/) (M A).7 y.77 (can earn at most 4/5) y 7.58 (allow 7.6) (A) Find differences (e.g. y ) and sample mean: d 6 /.6 M A Estimate population variance (to s.f.): s ( /) / 9 (allow biased here:.469 or.8 ).6 or.44 B State hypotheses (AEF; B for ), e.g.: H : µ y µ.4, H : µ y µ >.4 B Calculate value of t: t ( d.4)/(s/ ).64 M A State or use correct tabular t-value: t 9,.95.8[] B (or can compare d with.64) Consistent conclusion (AEF, ft on both t-values): [Accept H :] Wrong test can earn only B for hypotheses and B for conclusion No improvement of more than.4 B 5 8 Cambridge International Eaminations 5

36 Page 8 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Question Number Mark Scheme Details Part Mark Total A Find MI of disc about O: I disc ½ ma ma B Find MI of ring about O: I ring m (a) 8 ma B EITHER: Find MI of any rod about O: I rod ⅓ (m/)a (m/)(a) (/) ma B OR: Find MI of collinear rods about O: ⅓ (9m/)(a) ⅓ (m/)a ma (B) Find MI of object about O: I O ma 8 ma 4(/) ma 45 ma AG B Find MI of object about ais at O // to tangent: I O ½I O M Find MI of object about tangential ais at A: I A I O m (a) (5/) ma M A Find new MI when particle is attached at C: I A I A m (6a) (44/) ma M Find and use initial angular speed: ω u/a B Find gain in P.E. at instantaneous rest: (mg a mg 6a)( sin θ) M A 45mga/ 7mga/ or 48mga ( sin θ) 6mga A Find u by equating to rotational KE: ½ I A ω M u (6/44) 6ag [44 ] SR: Taking AC at sin (¼) to vertical: u (/7) (ag) or.7 (ag) A P.E. 48mga ( cos sin (¼)).54 mga (A) (ma 6/7) u.5[] (ag) (A) SR: Overlooking added particle can earn M B M A A M A (ma /7) B State suitable distribution: Geometric B State (at least) null hypothesis: (AEF) H : Distn. fits data or p.6 B (B for It is a good fit ) Find ep. values using pq with p.6, q.4: (ignore incorrect final value here, M e.g..495 which can earn ma 5/8 Combine last cells since ep. value < 5: O:... 6 E: B Calculate χ (result correct to s.f.): χ [] M A State or use consistent tabular value (to s.f.): 5 cells: χ 4, [or if or no cells combined: 6 cells: χ 5, cells: χ 6, or if 4 cells combined, as with 495: 4 cells: χ, ] B Valid method for reaching conclusion: Accept H if χ < tabular value M Conclusion (AEF, requires both values correct): 4. < 9.49 so distn fits or p.6 A (Allow A for It is a good fit ) Find prob. p of at least one 6 on 5 throws of one die: p (4 s.f.) M A Find prob. of at least one 6 on eactly 4 of dice: C 4 ; p 4 ( p) 6 M; M or.7 A Cambridge International Eaminations 5

37 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the May/June 5 series 9 FURTHER MATHEMATICS 9/ Paper, maimum raw mark This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the eamination. It shows the basis on which Eaminers were instructed to award marks. It does not indicate the details of the discussions that took place at an Eaminers meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Eaminer Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 5 series for most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademark of Cambridge International Eaminations.

38 Page Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Mark Scheme Notes Marks are of the following three types: M A B 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. 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 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: B or A means that the candidate can earn or. B// means that the candidate can earn anything from to. 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 s.f., or which would be correct to s.f. if rounded ( 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 9.8 instead of. Cambridge International Eaminations 5

39 Page Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 The following abbreviations may be used in a mark scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so etra 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 Approimation (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) Penalties MR PA A penalty of MR 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 " marks. MR is not applied when the candidate misreads his own figures this is regarded as an error in accuracy. An MR 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 approimation. The PA penalty is usually discussed at the meeting. Cambridge International Eaminations 5

40 Page 4 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Question Number Mark Scheme Details Part Mark Total Find T by equating dv/dt at t T to 6: 4T 4 6, T.5 M A Find radial component v /r of acceln. at t T: v /r (T 4T ) /.5 (M if T not given a value) (/) 4 [m s ] M A SR: Ma M (/4) if linear and angular confused 4 4 (i) Find ω from SHM eqn. d /dt ω at C:.65 ω, ω.65 or /6 B Find period T [s] from T π/ω: (ft on ω ) T π/¼ 8π (not 5.) B (ii) Find amplitude a [m] from v C ω (a ): 6 ω (a ) a 6 6, a M A Find time from C to M, e.g.: ω sin (/a) ω sin ½ or ω cos ( /a) ω cos ½ or ½T ω cos (/a) ω cos ½ M (AEF throughout) ω {.948 π/6 [.56]} or ω {.9656 π/ [.47]} or ω {π.76 π/ [.47]} A or or or 4.984;.67 [s] A; A Find v from conservation of energy: ½mv ½mu mga( cos θ ) M A Find R by using F ma radially: R mg cos θ mv /a B Eliminate v to find R: AG R mg( cos θ ) mu /a M A Find u or v in terms of cos θ when R : u ag( cos θ ) or v ag cos θ B EITHER: Replace cos θ in energy eqn with v u: 4u u ag ⅔(u ag) or u ag 8u M A OR: Find cos θ and substitute in energy eqn: [v /ag ] 4( cos θ ) cos θ cos θ 8/ 4u u ag ( 8/) (M A) Hence find u: u (ag/) or.46 (ag) A (i) Take moments for rod about B: W a cos W a cos (or with cos /) T a cos M A Hence find tension T: T 7W/ A (Can earn M A A if e.g. sin wrongly used) Find modulus λ using Hooke s Law: T λ (a a/5) / (a/5) λ (/7) (7W/) W/ M A 5 Cambridge International Eaminations 5

41 Page 5 Mark Scheme Syllabus Paper Cambridge International A Level May/June 5 9 Question Number Mark Scheme Details Part Mark Total (ii) EITHER: Find horizontal component of force F at B: X T cos (7 /4) W or. W B Find vertical component: (X, Y ft on T) Y 4W T sin 9W/4 B Find magnitude of F: F (X Y ) ( 57/) W or.77[5] W B Find direction of F (AEF): Upward force at angle to AB of tan Y/X tan /7 (A if direction unclear) 6.6 or.69 radians M A OR: Find component along CB: F (4W T) sin 5W/4 (B ) Find normal component: (F, F ft on T) F (4W T) cos ( /4)W (B ) Find magnitude of F: F ( 57/) W or 77[5] W (B) Find direction of F (AEF): Upward force at angle to CB of tan F /F tan /5 (A if direction unclear) 6.6 or.5 radians (M A) OR: Find component parallel to string CA: ±F T 4W sin W/ (B ) Find normal component: (F, F ft on T) ±F 4W cos W (B ) Find magnitude of F: F ( 57/) W or.77[5] W (B) Find direction of F (AEF): Upward force at angle to AC of tan F /F tan 4/ (A if direction unclear) 66.6 or.6 radians (M A) 5 5 For A & B use conservation of momentum, e.g.: mv A mv B mu M (m may be omitted here and below) Use Newton s law of restitution (consistent signs): v B v A eu M Combine to find v B : v B ( e) u/5 A For B & C use conservation of momentum, e.g.: mv B mv C mv B M Use Newton s law of restitution (consistent signs): v C v B e v B M Combine to find v C and v B : v C ( e ) v B / ( e) ( e ) u/5 AG A v B ( e ) v B / ( e) ( e ) u/5 A Find ratios or values of v A, v B, v C from momentum: v A v B v C [ u] B Find e from first collision eqns, e.g.: v A ( e) u/5 u/ (or find e and then use v A v B ) or v B ½ (u u) or (⅓ e) u ( e) u/5, e ⅔ M A Find e from second collision eqns, e.g.: v B v C so ( e ) ( e ) or v C ( ⅔) ( e ) u/5 u or v B ( ⅔) ( e ) u/5 u/ e ½ M A Equate pooled estimate of σ to : ( 5 /N 6 /) / (N ) M A Formulate and solve relevant quadratic eqn. for N: N 65 N 5, N 5 M A 4 4 Cambridge International Eaminations 5