Version.0 klm General Certificate of Education June 00 Mathematics MB Mechanics B Mark Scheme
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. Further copies of this Mark Scheme are available to download from the AQA Website: www.aqa.org.uk Copyright 00 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 (company number 364473) and a registered charity (registered charity number 073334). Registered address: AQA, Devas Street, Manchester 5 6EX
Key to mark scheme and abbreviations used in marking M m or dm A B 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 explanation or ft or F follow through from previous incorrect result MC mis-copy CAO correct answer only MR mis-read CSO correct solution only RA required accuracy AWFW anything which falls within FW further work AWRT anything which rounds to ISW ignore subsequent work ACF any correct form FIW from incorrect work AG answer given BOD given benefit of doubt SC special case WR work replaced by candidate OE or equivalent FB formulae book A, or (or 0) accuracy marks NOS not on scheme x EE deduct x marks for each error G graph NMS no method shown c candidate PI possibly implied sf significant figure(s) SCA 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 Examiner 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
MB (a) 30 seconds B B: Correct statement of time. (b) s = 40 0 400 = m A : A method for calculating the first distance. Must see 40 and. A: Correct distance. s = ( 0 + 0) 40 = 400 m () (A) 0 a = = 40 0 = 0 + s s = 0 = 400 m () (A) Note on third method: Must see or 0 plus attempt to find distance for 40. (c) s = 50 0 500 = m : Method for finding the second distance and calculating the total distance. s = ( 0 + 0) 50 = 500m () 0 a = = 50 5 0 = 0 + s 5 5 () Note on third method: Must see 5 plus attempt to find distance. or 0 50 s = 0 = 500 m 4 Total = 400 + 500 = 900 m AF AF: Correct total distance. Award the follow through mark for correct addition of 500 and their answer to (b). (d) 900 - v = = 7.5 ms AVERAGE 0 AF : Their total distance divided by 0 AF: Correct average speed based on their answer to (c). (e) 0 0 900 = 500 m AF : Multiplication of 0 and 0 to find distance. Note: Award if 400 seen in this part. AF: Correct difference based on their answer to (c) provided final answer is positive. Total 9 4
MB (cont) (a) B: Correct force diagram with arrows R or 98 or 0g and labels. F P Note: Award mark if forces drawn on the diagram in the question. Note: Do not accept 0 kg for the weight. mg or W or 0g or 98 Note: Do not accept μ R or 0.5R for F. or 9.8m B (b)(i) ( R = 0 9.8 = ) 98 N B B: Correct normal reaction. Accept 0g. No need to see the letter R or working. (ii) ( F )0.5 98 ( F )49 BF B: Correct maximum value for friction. Accept 5g. No need to see the letter F or any working. Ignore any inequalities. For FT, must be 0.5 of candidate s answer to (b)(i). (iii) ( F = )30N B B: Correct friction. Allow 30. (c) 80 49 = 0 a AF : Three term equation motion, containing 80, candidate s 49 and 0a (not 0ga) in any combination. AF: Correct equation including signs. a = 3.ms AF 3 AF: Correct acceleration. FT candidate s answer to (b)(ii). Total 7 Allow use of g = 9.8 (b)(i) 98. (b)(ii) 49.05 or 49. or 49 (c) 3.095 or 3.09 or 3. B B AA 5
MB (cont) 3(a) 3 7 6 + m = 6 + m : Four term conservation of 4 3 b A momentum equation. Allow sign errors. A: Correct equation with correct signs. Vector equation may be implied by later correct working in this part of the question. 6 + 3m = 6 + 7m A A: Correct equation for correct component. + 3m = 6 + 7 m 6 = 4m m =.5 A 4 A: Correct m. Example if only + 3m = 6 7 m without a vector equation award A0A0A0. (b) 6 4+.5 ( ) = 6 3+.5b BF BF: Correct equation using m or 4 3 = 8 +.5b 3=.5b b = BF Total 6 candidates m from (a). BF: Correct b from candidate s m from (a). 6 Note: b = m. Consistent use of mg instead of m throughout penalise mark. 6
50 cosθ = 60sin 4 (from vector triangle and sine rule) () (A) MB (cont) 4(a) 50 cosθ = 60 cos 48 A : Equation for two forces, with both forces resolved horizontally in the same way. (Accept 50sinθ = 60sin 48 for.) A: Correct equation. 50 = 60 sin 4 sin 90 θ (from Lami s Theorem) 50 = 60 sin38 sin 90 + θ For example: 60 cos 48 θ = cos 50 = 36.59 = 36.6 (to 3SF) (b) 50sin 36.59 + 60sin 48 = 9.8 m correct equivalent, for example: 50sin 36.59 + 60cos 4 = 9.8 m (from vector triangle and sine rule) 50 mg = sin 4 sin 84.6 (from Lami s Theorem) () (A) () (A) d A 4 AF () (AF) () (AF) (: Use of sine rule with 50, 60 and 4.) (A: Correct equation.) (: Use of Lami s Theorem with 50, 60 and 38.) (A: Correct equation.) d: Solving for θ. A: Correct θ. Note: Final answer of 63. from using resolving incorrectly with sines award A0dA0. Accept 36.5 (truncation) and 36.7 and AWRT 36.6. : Three term vertical equation, including mg with forces resolved vertically in the same way (accept 50cos 36.59 + 60cos 48 = 9.8 m for ). AF: Correct equation. (: Use of vector triangle and sine rule.) 50 60 = () sin38 sin 95.4 (AF) For example: 50sin 36.59 + 60sin 48 m = = 7.59 9.8 Total 7 (: Use of Lami s Theorem.) A 3 A: Correct value for m CAO. Accept 7.58, AWRT 7.6. Allow use of g = 9.8 (b) 7.58 AA 7
MB (cont) 5(a) A (b) ( v = ) 30 + 00 = 04.4 = 04 ms to 3SF 30 00 θ = tan 00 or tan 30 = 07 30 00 θ = sin 04.4 or sin 04.4 = 07 00 30 θ = cos or cos 04.4 04.4 = 07 A 3 AF () (AF) () (AF) Total 5 : Equation or expression to find v based on Pythagoras. Must be +. For example: 0 900 oe scores. A: Correct equation or expression, with square root. A: Correct v. Accept 04.4. : Trigonometric equation to find α. AF: Correct α. Follow through incorrect answer from (b). Note: Subtracting 7 etc from other values such as 360 or 90 can not be ignored and will score. Accept 6 or 7 or 6.6 or 6.7 or 6.8. Also accept all of these with a zero in front, eg 06. 8
MB (cont) 6(a) g T = a A : Three term equation of motion, with g (or 7.6), a (not ga) and T. A: Correct equation T 8g = 8a A : Three term equation of motion, with 8g (or 78.4), 8a (not 8ga) and T. A: Correct equation 4g = 0a 4 g a = =.96 ms 0 AG A 5 A: Correct acceleration from correct working. Note: Do not penalise candidates who consistently use signs in the opposite direction throughout, provided they give their final answer as.96. If final answer is.96 don t award final A mark. Special Case: Whole String Method 4 g = 0 a and 4 g a = =.96 OE AA 0 (b) T = 8 g + 8.96 = 94.N A : Use of three term equation of motion to find T, with a =.96. A: Correct tension. Accept 94.08. (c)(i) v = 0 +.96 = 3.9 ms A : Use of constant acceleration equation to find v, with a =.96 and u = 0. A: Correct v. Using s = 4 scores M0. (ii) v = 3.9 + 9.8 4 AF : Use of constant acceleration equation to find v, with a = ±9.8 and u 0. AF: Correct equation. FT initial velocity from (c)(i). v = 9.68 ms AF 3 AF: Correct v. FT initial velocity from (c)(i). For example.8 from 7.84. 9
MB (cont) (c)(iii) A 4 = ( 3.9 + 9.68) t : Use of s = A ( u+ v) t A: Correct values. A: Correct signs. t =.39 d d: Solving for t. A 5 A: Correct t. 4 = 3.9 t 4.9 t 4.9 t 3.9 t 4 = 0 3.9 ± 3.9 4 4.9 4 t = 4.9 t =.39 or t = 0.588 t =.39 t up + t = 0.4 + 0.4 + 0.588 down =.39 (to 3SF) 9.68 = 3.9 + 9.8t 3.6 t = =.39 9.8 () (A) (A) (d) (A) () (A) (d) (A) (A) () (A) (A) (d) (A) Total 7 : Forming a quadratic with candidates u from (c)(i) or v from (c)(ii)with 4.9 or 9.8. A: Correct terms in quadratic. A: Correct signs in quadratic. d: Solving quadratic (do not penalise for negative discriminant). A: Correct root seen (other root does not need to be seen). : Finding total time from two or three times. A: 0.4 or 0.8 seen. d: Finding second or third time for downward motion. A: Obtaining 0.588 or 0.988. A:.39. Accept.38. : Use of v = u+ at A: Correct values. A: Correct signs. d: Solving for t A: Correct t Use of g = 9.8 (a).96 AAA0 (b) 94. A (c) (ii) 9.69 AA (c) (iii).39 AAdA 0
MB(cont) 7(a) 0a = 9 i+ j : Application of Newton s second Law with m = 0 in vector form. a = ( 0.9 i+. j ) ms A A: Correct acceleration. If acceleration incorrect follow their value through for the rest of this question. (b)(i) r ( 5) = (. i+ j) 5 + ( 0.9 i+. j) 5 =.5 i+ 0 j d =.5 + 0 = 9.9 metres AF d AF 4 : Use of constant acceleration to find position vector at t = 5, with u 0 i+ 0 j. AF: Correct position vector, for candidate s acceleration which must be a vector. Allow.3i + 0j. d: Calculation of distance from position vector. Must see + sign. AF: Correct distance, for their acceleration. Accept 30 from.3i + 0j. (ii) = ( + ) + ( + ) v. i j 0.9 i. j t AF : Use of constant acceleration equation to find an expression for v, with u 0i+ 0j. AF: Correct v for their acceleration. (iii) = (. + 0.9t) + ( +.t) v i j. + 0.9 t = +. t. = 0.3t t = 4 AF AF 3 Total : Equation involving both i and j components of their velocity. Could have incorrect signs, for example. + 0.9t = ( +.t). AF: Correct equation. AF: Correct time, for their acceleration.
MB (cont) 8(a) 4.7sinα 9.8t ( = 0) A : Equation for vertical velocity being zero at highest point. Must have sinα with ± 9.8. A: Correct equation. 4.7sinα 3sinα t = = 9.8 AG A 3 A: Correct result from correct working. 4.7 sinαt 4.9T = 0 4.7 sinα T = = 3sinα 4.9 3sinα t = () (A) (A) All marks awarded for last line, from correct working. (b)(i) 3sinα 3sinα 7 = 4.7sinα 4.9 A 7 =.05sin α 7 α = sin = 5.8.05 d d A 5 : Expression including vertical displacement at height 7, using expression from part (a) and with ± g or equivalent. A: Correct expression. d: Simplified expression with sin α. d: Finding an angle. Must have previous d mark. A: Correct angle. Accept 5.7, 5.9. ( α ) 0 = 4.7sin + 9.8 7 9.8 7 sin α = 4.7 α = 5.8 () (A) (d) (d) (A) (ii) OA= 4.7cos5.8 3sin5.8 B B: Use of 3sinα with their α. : Finding horizontal displacement. including 4.7cosα with 3sinα or 3sinα. OA =. m A 3 A: Correct distance. Accept.3 m. (c) Ball is a particle/no spin. No air resistance/no wind/constant acceleration of 9.8/Only force is weight. B B Total 3 TOTAL 75 B: Particle assumption. B: Air resistance assumption. Use of g = 9.8: (a) AA0 (b)(i) 5.8 or 5.9 AddA (b)(ii). BA