A-LEVEL MATHEMATICS. Mechanics 2B MM2B Mark scheme June Version/Stage: Final V1.0

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1 A-LEVEL MATHEMATICS Mechanics B MMB Mark scheme 660 June 014 Version/Stage: Final V1.0

2 Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students scripts: alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this Mark Scheme are available from aqa.org.uk Copyright 014 AQA and its licensors. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre.

3 MARK SCHEME A-LEVEL MATHEMATICS MMB JUNE 14 Key to mark scheme abbreviations M mark is for method m or dm mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for explanation or ft or F follow through from previous incorrect result CAO correct answer only CSO correct solution only AWFW anything which falls within AWRT anything which rounds to ACF any correct form AG answer given SC special case OE or equivalent A,1 or 1 (or 0) accuracy marks x EE deduct x marks for each error NMS no method shown PI possibly implied SCA substantially correct approach c Candidate sf significant figure(s) dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded. of 10

4 MARK SCHEME A-LEVEL MATHEMATICS MMB JUNE 14 1 (a) KE J A1 (b) Change in PE; mgh J A1 8 J SC1 80 (c)(i) Salmon s KE when it reaches the sea (a) + (b) J [both non zero] 478 J A1 Ft [one correct] (ii) Speed of salmon is ms ms -1 A1 Accept 17.8,17.85,17.855, Total 8 4 of 10

5 MARK SCHEME A-LEVEL MATHEMATICS MMB JUNE 14 (a) Using F ma either term correct oe a 4e t i t j A1 (b) v adt for either term correct e t i 1 t4 j + c A1 Ft from (a) oe Condone no + c When t 0, 7i 4j i + c m1 c 5 i 4 j v ( e t + 5)i ( 1 t4 + 4)j A1 4 CAO (c) When t 0.5, v ( e 1 + 5)i ( )j A i j Speed is A1 4 or 7.01 ms -1 Total 10 MR A0 in (a) and last part of (c) Do not accept 7 X for at least correct or 6. A1 14 Accept 0 Y or 6.15 A1 0 Centre of mass is at (6., 6.15) A1ft 5 Do not accept 0 (6.15,6.)MA If lamina not used SC; ie, Total 5 14 etc 5 of 10

6 MARK SCHEME A-LEVEL MATHEMATICS MMB JUNE 14 4 (a) 0 revolutions per minute 40π radians per minute B1 or 1 revolutions per second π radians per second B1 Accept.09 (b) Resolve vertically Tcos 5 0.8g A1 if Tsin5 used; need g T A N (c) Resolve horizontally T sin 5 mω r condone Tcos5 and m v r π 9.57 sin r ( ) A1 A1 A1 for either side r A1 4 Radius is 1.56 m Condone 1.57 Total 9 5 (a) Using conservation of energy : 1 1 mvp mvq + amg for [or 4] terms A1 KE and 1[or ] PE v Q 49ag 4ag vq 45ag v Q 45ag A1 4 v Q 5ag (b) At Q, T + mg mv Q a A1 for correct terms T m. 45 g mg 44mg A1 Total 7 6 of 10

7 MARK SCHEME A-LEVEL MATHEMATICS MMB JUNE 14 6 (a) Using F ma 1 0.mv m dv 1 dv dt 0.v dt B1 Need substitution for a 1 v dv 0.dt v 0. t + c When t 0, v 8, c 6 v 0. t + 6 v 0. t + 4 v ( 4 0.t) A1A1 A1 A1 6 (b) When v 0, 4 0. t 0 t 0 A1 A1 for each side no sign [B0] could get A1 (c) Integrating v ( 4 0.t), 5 x (4 0.t) + d A1 When t 0, x 0, d 64 5 A1 x (4 0.t) + 64 When speed is 0 ms 1, t 0 x 64 A1 5 Total 1 for power of 5/ A1 correct [condone no d] 7 of 10

8 MARK SCHEME A-LEVEL MATHEMATICS MMB JUNE 14 7(a) B Need 5 forces correct B S ignore labels C D B1 for 4 forces correct 88g g 4m 60 R F A (b) Resolve horizontally F S cos 0 Resolve vertically R 88g + g S sin 0 Moments about A g. cos g. 4 cos 60 5 S 5S 09g B1 B1 for correct moments about any point S 41.8g [409.64] A1 Using F µr; S cos 0 µ(110g S sin 0) µ S 0g S A1 6 Resolve once B1 moments twice is A1, B1 R F Accept 0.407, ,0.41 not 0.4 If S is horizontal, B1 in (a) In (b) [moments], for friction,b1 [ resolve] 0.49 SC Total 8 8 of 10

9 MARK SCHEME A-LEVEL MATHEMATICS MMB JUNE 14 9 of 10

10 MARK SCHEME A-LEVEL MATHEMATICS MMB JUNE 14 8 (a) Resolve perpendicular to plane R mg cos 0 F μr μmg cos0 m gcos 0 or 0.8 x A1 9.5 N (b)(i) As particle moves from C to B; Constant friction acts. Work done by friction is (x + ) Change in PE is mg(x+)sin 0 λ x Initial EPE l 10 ( x 1.5) 1. 5 B1 B1 40 ( x 1.5) B1 10 (0.5) Final EPE B1 (x + ) mg(x+)sin 0 A1 40 ( x 1.5) 10 A1 for 4 of these terms at least correct A1for terms correct with correct signs A1 for equation totally correct 40 x x x or x 4.11 A1 8 condone 4.10, 4.1, and anything in between, (ii) λ Using T lx Tension when particle is at B is B1 Frictional force is Gravitational force is mg sin B1 For both 9.4. and 1.4. Using F ma 4a Need all terms & correct.98 Acceleration is ms - A1 4 condone 5.99, ,5.985 Total 15 TOTAL of 10

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