Version.0 klm General Certificate of Education June 00 Mathematics MM03 Mechanics 3 Mark Scheme
Mark schemes are prepared by the Principal Eaminer 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 eaminers and is the scheme which was used by them in this eamination. The standardisation meeting ensures that the mark scheme covers the candidates responses to questions and that every eaminer understands and applies it in the same correct way. s preparation for the standardisation meeting each eaminer 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, eaminers encounter unusual answers which have not been discussed at the meeting they are required to refer these to the Principal Eaminer. It must be stressed that a mark scheme is a working document, in many cases further developed and epanded on the basis of candidates reactions to a particular paper. ssumptions 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 eamination paper. Further copies of this Mark Scheme are available to download from the Q Website: www.aqa.org.uk Copyright 00 Q and its licensors. ll rights reserved. COPYRIGHT Q retains the copyright on all its publications. However, registered centres for Q are permitted to copy material from this booklet for their own internal use, with the following important eception: Q 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 ssessment and Qualifications lliance. The ssessment and Qualifications lliance (Q) is a company limited by guarantee registered in England and Wales (company number 364473) and a registered charity (registered charity number 073334). Registered address: Q, Devas Street, Manchester 6EX
MM03 - Q GCE Mark Scheme 00 June series Key to mark scheme and abbreviations used in marking M m or dm E mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for eplanation or ft or F follow through from previous incorrect result MC mis-copy CO correct answer only MR mis-read CSO correct solution only R required accuracy WFW anything which falls within FW further work WRT anything which rounds to ISW ignore subsequent work CF any correct form FIW from incorrect work G answer given OD given benefit of doubt SC special case WR work replaced by candidate OE or equivalent F formulae book, or (or 0) accuracy marks NOS not on scheme EE deduct marks for each error G graph NMS no method shown c candidate PI possibly implied sf significant figure(s) SC substantially correct approach 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. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Eaminer will alert you to these and details will be provided on the mark scheme. 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. 3
MM03 - Q GCE Mark Scheme 00 June series MM03 LT For dimensions of u α β γ 3 3 LT = M L T L ML LT for equation with five components = β + = γ 0= α + m β = 0, α = -, γ = The dimensions of C are M T F Forming and solving equations (PI) lternative : LT () For dimensions of u 3 3 LT = C L ML LT () for equation with five components LT = C LMT (m) The dimensions of C are M T (F) Total (a)(i) = 80cos θ.t t = 80cosθ y = 80sinθ.t gt y = 80sinθ g( ) 80cos θ 80cos θ y tanθ ( tan θ ) g = + 800 nswer given (ii) 9 8 400 0 = 400 tanθ. ( + tan θ ) 800. tan θ 400tanθ + 0. = 0 49 tan θ 60 tanθ 4 0 Condone + 0 + = nswer given (b)(i) 60 ± 600 4( 49 )( 4) tanθ = 49 =.980, 0.803 PI θ = 7.,.7 F 3 (ii) For the shortest time 400 = 80cos.7.t t = 9. IF (c) The projectile is a particle The air resistance is negligible E Total 3 4
MM03 - Q GCE Mark Scheme 00 June series MM03 (cont) 3(a) C.L.M. ()3 u = () v + (3) v for three non-zero terms Restitution : 3 u = v v 3 ccept v v v = u m Solution v = 0 6 for both answers (b) C.L.M. 3u = 3w + w c Restitution : u = wc w 3 4u m Solution attempt, dep. on both s wc = 3 + G u( 9 ) w = OE 33 ( + ) 6 for both (c) u( 9 ) For further collision < 0 33 ( + ) 9 u u < 0 > 9 G (d) I = ( 4u ) 3 + I = u lternative: u( 9 ) I = 3u 3 33 ( + ) () I = u (F) Total 6 u ccept Follow through on their w
MM03 - Q GCE Mark Scheme 00 June series MM03 (cont) 4(a) r = ( 60i+ 30 k) + (0i+ 0j 00 k ) t For correct form r = ( 40i + 0j 0 k) + (00i+ j + 0 k) t, 3 for each (b) [ ( 60i 30k) ( 0i 0j 00k) ] r = + + + t ttempt at the difference using their answers [( 40i+ 0j 0 k) + (00i+ j+ 0 k )t] r = ( 0 + 0 t ) i+ ( 0 + t ) j+ (40 0 t ) k G (c) For collision ( 0 + 0 t) i+ ( 0 + t) j+ (40 0 t) k = 0 0 + 0t = 0 t = 0 + t = 0 t = m F 40 0t = 0 t = 4 The relative position vector cannot be zero. Therefore and do not collide E 4 (d) S = ( 0+ 0 t) + ( 0+ t) + (40 0 t) For minimum S ds = 00( 0+ 0t) + 0( 0+ t) dt 300( 40 0t) = 0 F 0t 400 = 0 m Solution t = 0. 83 F 6 Total lternative: 0 + 0t 0 0 t. () + = 0 40 0t 0 () 000 + 00t 0 + 6t 6000 + 00t = 0 (m) (F) 6t 70 = 0 (F) t = 0. 83 (F) 6
MM03 - Q GCE Mark Scheme 00 June series MM03 (cont) (a) Parallel to the wall 4cosα = vcos40 Correct trigonometric ratios Perpendicular to the wall vsin40 = 4sinα 3 Correct trigonometric ratios tan 3 α = tan 40 3 G (b) α =. 4cos. v = cos40 - v = 3. ms 3 OE Total 6 6(a) The spheres are smooth, no force acting in j direction E ny valid reason (b) v = ai + bj v = ci + dj C.L.M. along i : () + ( ) = ( a) + ( c) a+ c= 0 Restitution along i : c a = 0. ( ( )) c a =. c = 0. a = v = i + 3j F v = 0. i j F 6 Total 7 7
MM03 - Q GCE Mark Scheme 00 June series MM03 (cont) On striking : 7(a) 0sin30.t (9. 8)cos3.t = 0 t = 49. WRT OE Components of Velocity : u = 0cos30 9. 8sin3 (. 49) u = 33. F WRT u = 0sin30 9. 8cos3 (. 49) y u = 0 (or -9.99) F 7 y (b) On Rebounding v = 33. v 4 y = 0 F For 4 their v = 8 (or 7.99) y The rebound angle = tan 33. = 67. ( or 67. 4 ) F 8 3 + 67. = 0. F 0. > 90, therefore the second strike will be at a point lower down than. E 6 Dependent on the two s lternative: 4 0 8 = () Condone negative sign 0= 8t g cos3.t () t = 993. () OE = 33. t g sin3.t () = 4. or -4.6 () The second strike will be at a point lower down than. (E) Total 3 TOTL 7 u y 8