Mark Scheme (Results) January 2012 GCE Mechanics M3 ... - Edexcel

Mark Scheme (Results)January 2012GCE Mechanics M3 (6679) Paper 1

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General Marking Guidance• All candidates must receive the same treatment. Examiners must mark thefirst candidate in exactly the same way as they mark the last.• Mark schemes should be applied positively. Candidates must be rewarded forwhat they have shown they can do rather than penalised for omissions.• Examiners should mark according to the mark scheme not according to theirperception of where the grade boundaries may lie.• There is no ceiling on achievement. All marks on the mark scheme should beused appropriately.• All the marks on the mark scheme are designed to be awarded. Examinersshould always award full marks if deserved, i.e. if the answer matches themark scheme. Examiners should also be prepared to award zero marks if thecandidate’s response is not worthy of credit according to the mark scheme.• Where some judgement is required, mark schemes will provide the principlesby which marks will be awarded and exemplification may be limited.• When examiners are in doubt regarding the application of the mark schemeto a candidate’s response, the team leader must be consulted.• Crossed out work should be marked UNLESS the candidate has replaced itwith an alternative response.

General Principals for Core Mathematics Marking(But note that specific mark schemes may sometimes override these general principles).Method mark for solving 3 term quadratic:1. Factorisation( x 2 + bx + c)= ( x + p)(x + q),where pq = c, leading to x = ....2( ax + bx + c)= ( mx + p)(nx + q),where pq = c and mn = a, leading to x = …2. FormulaAttempt to use correct formula (with values for a, b and c), leading to x = …3. Completing the square2Solving + bx + c = 0x b: ( ) 2x ±2± q ± c, q ≠ 0 , leading to x = …Method marks for differentiation and integration:1. DifferentiationPower of at least one term decreased by 1. (n →2. IntegrationPower of at least one term increased by 1. (n − 1x x )+ 1x n → xn)Use of a formulaWhere a method involves using a formula that has been learnt, the advice given in recentexaminers’ reports is that the formula should be quoted first.Normal marking procedure is as follows:Method mark for quoting a correct formula and attempting to use it, even if there are mistakesin the substitution of values.Where the formula is not quoted, the method mark can be gained by implication from correctworking with values, but may be lost if there is any mistake in the working.

January 20126679 Mechanics M3Mark SchemeQuestionNumberSchemeMarks1.λ × 0.52EPE =1.2B1GPE lost = EPE gainedM1 (used)2λ × 0.50.8×9.8×1.1 =1.2A1ftλ = 41.4 N or 41 N A14

QuestionNumberSchemeMarks2.(a)2π2πT = = ,ω 3ω = 3B12a = ω x = 9×0.2 = 1.8 ms -2 towards C M1 A1v = ω a xM1v = 1.37 ms -1A12 2 2 2(b) ( − ) = 9(0.25 − 0.04) = 1. 89(c) x = 0 .5sin 3t= 0. 2M1 A1ft1 1−t = sin 0.4 ≈ 0.137 s A13(3)(2)(3)8

QuestionNumber3.(a)(b)dv10 10 100a v − −= = × , =d x x + 6 x + 6 x + 6( ) ( )2 3dx10= ⇒ ∫ x + 6dx=+∫10dtdtx 614=( )3Scheme−100 = −114 + 6 80ms -2MarksM1M1, A1A1M1M12⎡ x ⎤T⎢+ 6x= [ 10t]21⎣⎥⎦2M1 A1196+ 6 × 14 − 2 − 12 = 10 T − 102M1178 = T T = 17.8(s)A1(4)(6)10

QuestionNumberSchemeMarks4.(a)A60°TrB0.5g(b)↑ T cos 60° = 0. 5g, T = g (1) M1, A1x 19.6xExtension in the string = x, T = λ =a 0.8B1Using (1) , g = 24.5x,x = 0.4 m * M1, A12→ T sin 60° = 0.5×r × ω (2) M1 A1Using (2)2g sin 60°= 0.5×(0.8 + 0.4)sin 60°ωM1 A12 2g5gω = , ω = (4.04 or 4.0)1.2 3A1(5)(5)10

QuestionNumberSchemeMarks5.(a)(b)Distance of P from the centre of the Earth = R + xkF =R + x( ) 2x = 0, F =mg, mg( R) 2( R + x) 2k = M1 A12mgRF = * A1F = ma ,2gR dv− = vM1 A12( R + x) dxgR2 2R2gRvdv= − dxM1 A1∫ ∫ 2VRgR2( R + x)2R2⎡12 ⎤ ⎡ gR ⎤M1⎢v ⎢ ⎥2 ⎥=⎣ ⎦V⎣ R + xA1⎦ R2 21 gR 1 2 gR gR gR× − V = − = −2 2 2 3R2R6M12V gR gR 5gR 2 5gR5gR= + = V = , V =2 4 6 12 66A1, A1(3)(9)12

QuestionNumber6.(a)SchemeMarksl(b)(c)GPE gained = mgl ( 1− cosθ )11gl1 11gl1 2 4Conservation of energy: m = mgl( 1−cosθ ) + mv2 42M1A1 A12 ⎛11⎞ ⎛ 3 ⎞v = gl⎜− 2 + 2cosθ⎟ = gl⎜+ 2cosθ⎟⎝ 4⎠ ⎝ 4 ⎠Resolving towards the centre of the circle:M12mvT − mg cos θ =lA1 A1⎛ 3 ⎞T − mg cosθ= mg⎜+ 2cosθ⎟⎝ 4 ⎠M1⎛ 3 ⎞ ⎛ 1 ⎞T = mg⎜+ 3cosθ⎟ = 3mg⎜cosθ+ ⎟ *⎝ 4 ⎠ ⎝ 4 ⎠A1(8)1T = 0 ⇒ cosθ= −4M12 ⎛ 3 ⎞ gl glv = gl⎜+ 2cosθ ⎟ = , v =⎝ 4 ⎠ 4 4M1, A1(3)Horizontal component of velocity at B=gl1 gl× cos( 180 −θ ) =44 4B1ft1 gl 1 glExtra height h ⇒ mgh + m = m2 64 2 4M1 A1⎛ 1 1 ⎞ 15lh = ⎜ − ⎟l= (0.117l)⎝ 8 128 ⎠ 128A1gl 15sin 2 ×v θ 15OR: Using4 16 lh = = =2g2g128(4)2 2 15gl15lOR: Using v = u + 2as,0 = − 2gh,h =6412815

QuestionNumber7.2 2(a) y dx∫ x ( 6 − x) dx= ∫Scheme2 ππ 2 3 4∫ π = 36x−12x+ x dxM1 A1445π ⎡3 4 x ⎤ π 1024= ⎢12x − 3x+ ⎥ = ×M14 ⎣5 ⎦ 4 52(160.8….)2 3( ) 23 4 5π y xdx =πx 6 − x dx =π36x − 12x + x dx4 4∫ ∫ ∫ M1 A1π ⎡ 4 12 5 1 6 ⎤ π 10496= 94 ⎢ x − x + x = ×⎣ 5 6 ⎥⎦ 24 15M1(549.5……)10496 5⇒ x = × = 3.416......15 1024M1 A1Required distance ≈ 3.42 – 2 = 1.42(cm) *A1(b)1Base has radius × 2 × 4 = 4 cm2B14About to topple ⇒ tan α =1.42M1 A1α ≈ 70. 5°A1(c) Parallel to slope: F = mg sin βPerpendicular to the slope: R = mg cos βM1 A1About to slip: F = µ Rtan β = µ = 0.3, β ≈ 16. 7°A166Marks(9)(4)(3)16

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