PMT A-LEVEL MATHEMATICS. Mechanics 1B MM1B Mark scheme June Version/Stage V1.0 Final

Transcription

1 A-LEVEL MATHEMATICS Mechanics 1B MM1B Mark scheme 660 June 014 Version/Stage V1.0 Final

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

3 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.

4 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. 4 of 14

5 No MR or MC in this question. 1(a) v = = -1.6 m s M1: Using constant acceleration equation(s) to find v. : Correct equation for v. : Correct answer. CAO NMS Full marks for correct answer. Do not accept 4. (b) 1 s = = 5.6 = 6 m (to sf) 1 s = (14 +.6) 1 = 5.6 = 6 m (to sf) 1 s = = 5.6 = 6 m (to sf).6 = s s = 5.6 = 6 m (to sf) (F) () (F) () (F) () () () () M1: Using a constant acceleration equation or area under a velocitytime graph to find s. (F): Correct equation using their answer to part (a) if their equation contains v. : Correct answer. Accept 5.6 or 6 Do not accept 5 unless 5.6 seen first. Do not accept 0. (c) 1600 R = R = 480 N M1: Three term equation of motion with correct terms. Allow sign errors. : Correct equation. : Correct R. CAO NMS Full marks for correct answer. Total 9 5 of 14

6 No MR or MC in this question. (a) M1 M1: Equation or expression to find F or F = F based on Pythagoras. Must have a +. = 500 : Correct force. Accept AWRT 7.1. = 7.1 N Alternative if angle found first. 40 F = = 7.1 sin.69 F = 60 = 7.1 cos.69 (M1) () (M1) () Accept 0 1. Do not accept 7. Note that just F = Scores M1A0. M1:Using 40 and sin of their angle 60 and cos of their angle. : Correct force. Accept AWRT 7.1. Do not accept 7. (b) ( F sinθ = 40) ( F cosθ = 60) F sinθ 40 = tanθ = F cosθ 60 θ =.7 M1: Seeing tan with 40 and 60. can be upside down. : Correct expression for tan θ. : Correct angle. Accept.6 or AWRT.7. Or 40 sinθ = 500 θ =.7 Or 60 cosθ = 500 θ =.7 (M1) (F) () (M1) (F) () () () Do not accept 4. M1: Use of sin with 40 or 60 in the numerator and their answer to (a) as the denominator. F: Use of sin and 40 in numerator, provided expression satisfies 1 sin θ 1. : Correct angle. Accept.6 or AWRT.7. M1: Use of cos with 40 or 60 in the numerator and their answer to (a) as the denominator. F: Use of cos and 60 in the numerator, provided expression satisfies 1 cos θ 1. : Correct angle. Note that a final answer of 56. scores M1A0A0. Total 5 Accept answers in radians (0.588). 6 of 14

7 No MR or MC in this question. (a) v = M1: Using constant acceleration equation(s) to find v, with u = v = m s : Correct equation. : Correct answer. Accept or AWRT or 15 5 (b) F or µr 1 or 0.4 R 1 or 16 R 1 or R or N 1 T mg or 800g or W 1 or 7840 Do not accept 0.77 unless AWRT seen first. Note that if using two equations the time is 15.49s. B1: Correct diagram with exactly four forces showing arrow heads and labelled. Note: Do not accept 800 kg for the weight. B1 1 (c) R or S or N T P Mg or 1700g or W or B1 B1: Correct diagram with exactly four forces showing arrow heads and labelled. Do not award if a different letter is used for the tension in this part. Condone the use of the same letter for the normal reaction and the weight / mass in both parts. Condone assumption that normal reaction is equal to the weight when labelling diagrams. Condone diagrams with two or four normal reactions if they are seen acting at the wheels. Note: Do not accept 1700 kg for the weight. Only one normal reaction force / Don t need to consider where the normal reaction forces act. B1 B1: Any comment about the number of normal reaction forces or where they act. (d) R = M1 ( = 7840 N) F = dm1 = 16 = 140 N (to sf) M1: Correct reaction force. dm1: Use of F = µ R. Accept 800g for R. Condone negative signs in working. : Correct friction. Only accept 140 or 16. Do not award for negative final answers. 7 of 14

8 (e) T 16 = F T = 176 = 180 N (to sf) M1: Three term equation of motion for the skip with correct terms. Allow sign errors and their value of F from part (d). Must use mass of 800 kg. F: Correct equation with their value for F. : Correct T. Only accept 176 or 180. (f) P 176 = P = 61 = 60 N (to sf) P 16 = P = 61 = 60 N (to sf) F (F) () () M1: Three term equation of motion for the van with correct terms. Allow sign errors and their value of T from part (e). Must use mass of 1700 kg. F: Correct equation. Follow through their value for T. : Correct P. Accept 61. Allow 65 or 60 or 70 from use of 180 for T. M1: Three term equation of motion for the van and skip with correct terms. Allow sign errors and their value of F from part (d). Must use mass of 500 kg. F: Correct equation. Follow through their value for F. : Correct P. Accept 61. Allow 65 or 60 or 70 from use of 140 for F. Total 15 Use of g = 9.81 gives: (d) 19. (e) 179. (f) 64. Do not penalise use of of 14

9 No MR or MC in this question. 4(a) 0 M1: Use of distance divided by speed (0 t = sin 70 sin70 or cos70 or sin0 or cos0 = 10.6 seconds sin ). sinα Value forα does not have to be seen. : Correct expression for t, with α = 81.5 if included. : Correct time. Accept AWRT (b) 0 tanα = α = 81.5 M1 M1: Use of 0 tan α = or in the. Allow use of sin or cos with denominator. : Correct angle. Allow AWRT 81.5 or 81, but not (c) = 8.5 V = sin 8.5 sin sin 8.5 V = = sin ( V cos70 ) = V = d () () (d) 5 M1: Forming an equation to find V., : Correct equation award A, if one error award. dm1: Solving equation for V : Correct V. Accept AWRT 0.97 from correct working, particularly from working with 10.6 or 8.5 etc. 0 tan 0 + = 10.64V V = () () (d) V = cos70 + sin 70 tan 8.5 = () () (d) Total 10 9 of 14

10 Follow through mis-reads in this question. 5 m(4i + j) + km(6i j) = ( m + km)(5.i 0.4j) 4m + 6km = 5.( m + km) 4 + 6k = k 0.8k = 1. k = = 1.5 m km = 0.4( m + km) k = k.4 = 1.6k k = = 1.5 B1 dm1 () (dm1) () 5 B1: Correct vector equation for conservation of momentum. Can be implied in later working by correct equation for one component. Condone missing brackets if later working correct. M1: Conservation of momentum equation for one component with correct terms, but allow sign errors. : Correct equation. Allow inclusion of i or j in equations. dm1: Valid method for solving for k. Working like 1.i k = = i scores dm0. : Correct value for k. CAO Candidates who work with both components and get two different values for k, should be treated as two solutions. (Mark both and take mean, rounding down.) 1.i +.4j = k( 0.8i + 1.6j) = k ( ) 7. k =. k = ± 1.5 k = 1.5 () (dm1) () Total 5 M1: Forming equation for k using magnitudes of vectors. : Correct equation. dm1: Obtaining ±1.5 : Selecting 1.5. Consistent use of weight instead of mass deduct 1 mark. 10 of 14

11 No MR or MC in this question. 6(a) V cos 1.8 = 40 M1: Equation to find V with = V 1.8 (cos or sin ). V = =.47 = m s (to sf) 1.8cos : Correct equation. : Correct V. Allow.5 or AWRT from correct working. (b) 1 y =.47sin = 1.09 = 1.1m (to sf) Distance between A and C is 1.1 metres. F 4 M1: Use of constant acceleration equation to find y (vertical displacement). Allow cos and ±g. F: Correct equation. Follow through their V from part (a). : Correct y. Accept AWRT Accept-1. or -1.4 from use of, provided correct working seen in part (a). Accept-1.0 or -1.1 from use of.5, provided correct working seen in part (a). : Correct distance. Accept modulus of answers as above. (c) No air resistance. B1 1 B1: No (air) resistance or only force is weight. Do not allow only force acting is g. Total 8 Use of g = 9.81 gives: (b) Use of gives Use of.5 gives Do not penalise use of of 14

12 Candidates who use a = 0.4i + 0. jfor both particles can gain all method marks but no A or B marks. Otherwise no MR or MC in this question. 7(a) 1 1 r = (4i + j) 10 + ( 0.4i) 10 M1 M1: Use of ut+ at with a = 0i + 0j non-zero acceleration. : Correct position vector. (b) r = (4i + j) t 0.t i B1 B1: Seeing position vector for A A, may be implied by later rb = (0.4i + 0.6j) t + 0.1t j+ 11.i B1 working if not seen explicitly. B1: Seeing position vector for (4i + j) t 0.t i = (0.4i + 0.6j) t + 0.1t j+ 11.i B, may be implied by later t = 0.6t + 0.1t working if not seen explicitly. M1: Forming an equation by t 14t = 0 equating either i or j ( t = 0) or t = 14 components to form a quadratic equation. 4t 0.t = 0.4t d : Correct equation t 18t + 56 = 0 : Correct solution(s). dm1: Forming an equation by t = 4 or t = 14 equating other components to t = 14 form a quadratic equation. : Correct equation : Correct solutions and selection of common solution. THEN = = 16.8 t = 14 r = (4i+ j) i= 16.8i+ 8j (dm1) () () B1 9 Total 11 (dm1): Substituting one nonzero value from their two solutions into the other component for both A and B. (): Obtaining two identical values. ():Concluding that t = 14. B1: Correct position vector. 1 of 14

13 No MR or MC in this question. 8(a) F = sin 0 = 196 R = cos0 = sin 0 = µ cos0 sin 0 µ = = tan 0 = cos0 µ = = or F = mg sin 0 or R = mg cos0 dm1 5 M1: Resolving parallel and perpendicular to the slope to obtain expressions for F and R. Allow consistent mixing of sin and cos. : Correct F. : Correct R. dm1: Use of F = µ R : Correct µ. Accept or AWRT or AWRT NMS gives full marks. 1 or (b)(i) (b)(ii) R = X sin cos0 ( = 0.5X + 9.5) X cos sin 0 F = X cos0 196 tan 0 (0.5X + 9cos0 ) = cos0 tan 0 X = = 69.8 = 69 N (to sf) cos0 0.5 tan 0 Alternative format dm1 dm1 5 Complete solution with omission of g throughout with correct final answer scores SC4. M1: Resolving perpendicular to the plane with three terms; R, Xsin0 or Xcos0 and 40gcos0 or 40gsin0, with consistent mixing of sin and cos. Allow sign errors. : Correct expression for R, may include m and g. M1: Equation of motion with correct terms. Allow sign errors. : Correct equation. May be in terms of m and g. dm1: Substituting for F using F = µr, where R is in the form a + bx where a and b are nonzero constants. dm1: Solving for X. : Correct value for X. Accept 69 or AWRT 69. X cos0 40g sin 0 F = X cos0 40g sin 0 tan 0 (0.5X + 40g cos0 ) = g sin g cos0 tan 0 X = = 69.8 = 69 N (to sf) cos0 0.5tan 0 Total 1 TOTAL 75 Use of g = 9.81 gives: (b) (i) 0.5X (b) (ii) 69.5 = 694 (to sf) Do not penalise use of of 14

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