abc GCE 2005 Mark Scheme January Series Mathematics MM1B

Transcription

1 GCE 005 January Series abc Mark Scheme Mathematics MB Mark schemes are prepared by the Principal Examiner and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation meeting attended by all examiners and is the scheme which was used by them in this examination. The standardisation meeting ensures that the mark scheme covers the candidates responses to questions and that every examiner understands and applies it in the same correct way. As preparation for the standardisation meeting each examiner analyses a number of candidates scripts: alternative answers not already covered by the mark scheme are discussed at the meeting and legislated for. If, after this meeting, examiners encounter unusual answers which have not been discussed at the meeting they are required to refer these to the Principal Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of candidates 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. Copyright 005 AQA and its licensors

2 Further copies of this Mark Scheme are available to download from the AQA Website: Copyright 005 AQA and its licensors. All rights reserved. COPYRIGHT AQA retains the copyright on all its publications. However, registered centres 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 centres to photocopy any material that is acknowledged to a third party even for internal use within the centre. Set and published by the Assessment and Qualifications Alliance. The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales and a registered charity number Registered address AQA, Devas Street, Manchester. 5 6EX. Dr Michael Cresswell Director G

3 Mark Scheme Mathematics Advanced Subsidiary and Advanced MB 1(a)(i) 40 = a Use of a constant acceleration equation to 40 1 form equation for a - a = = 0.8 ms AG AG; correct answer from correct working 100 (ii) 1 s = (1 + 40) 100 = 600 m Expression for distance, using t =100 Correct final distance (c) F = F = = N (a) = v v = = ms Three term equation of motion Correct force Three term momentum equation Solving for v 3 (a) u = u = = ms 3 Three term momentum equation F T B1 1 Correct diagram mg 40cos30 F = F = 40cos30.5 = 3.1 N (c) R + 40sin 30 = R = 5 N 3 3 Three term equation of motion AG; correct force from correct working Resolving vertically AG; correct force from correct working (d) 3.1 = 5µ 3.1 µ = = (e) Friction will decrease as normal reaction B1 decreases B1 Total 11 use of F = Correct µ µ R Decrease in friction Normal reaction decreases Copyright 005 AQA and its licensors 3

4 Mathematics - Advanced Subsidiary and Advanced Mark Scheme MB (cont) 4(a) Light or smooth B1 1 Acceptable assumption 5g T = 5a T g = a 3g = a 3g a = = 4. ms - (c) T = = 8 N 5 Total 8 5(a) 00sin 30 = T sin 45 00sin 30 T = = 141 N sin cos cos 45 + R = R = 4630 N 4 Three term equation of motion for one particle Three term equation of motion for other particle AG; correct acceleration from correct working Substitute a = 4. into one equation of motion Correct tension Resolving horizontally AG; correct T from correct working Resolving vertically with four terms Correct values Correct signs Correct R 6(a) sin 60 sinα = 6 α = Use of sine rule Correct LHS Correct RHS AG; correct α from correct working v = sin( ) v = 6.4 or 6.5 ms 6 sin 60 3 use of sine rule to find v Correct v 4 Copyright 005 AQA and its licensors

5 Mark Scheme Mathematics - Advanced Subsidiary and Advanced MB (cont) (a) i + j = i j + 10a Use of velocity equation a = 0.3i + 0.j 3 Correct a 1 r = ( i j) t+ ( 0.3i+ 0. j) t + 0i = t 0.15t + 0 i + t+ 0.1t j ( ) ( ) (c) (i) r(0) = ( ) i+ ( ) j = 0i+ 0j so due north of origin ft 3 Use of constant acceleration equation for position Correct i component Correct j component ft incorrect acceleration Substituting t = 0 into their expression for r Correct conclusion from correct working (c)(ii) v(0) = i j+ 0( 0.3i+ 0. j) = 4i+ 3j v (0) = = 5 ms 8(a) Ball is a particle No air resistance 1 (i) 0 = 1sin t 1sin 40 t = = 0.8 s 9.8 ft 4 Total 1 B1 B1 4 Finding velocity at t = 0 Finding magnitude Correct speed ft incorrect acceleration One appropriate assumption Second appropriate assumption Equation to find time at maximum height Solving for t Correct time (ii) h = 1sin = 3.04 m 3 Substituting time from previous into expression for height Correct expression AG; correct height from correct working (c).44 = 1sin 40 t 4.9t + = 4.9t 1sin 40 t.44 0 t = or s = 1cos = 10.4 m Total 16 TOTAL 5 Equation for time to get to the bar, based on height being.44 Correct LHS Correct RHS Solving quadratic Correct time / times Substituting their larger time into an expression for the horizontal displacement Correct distance Copyright 005 AQA and its licensors 5