2006 Applied Mathematics. Advanced Higher Mechanics. Finalised Marking Instructions

Transcription

1 006 Applied Mathematics Advanced Higher Mechanics Finalised Marking Instructions The Scottish Qualifications Authority 006 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is to be used for any other purposes written permission must be obtained from the Assessment Materials Team, Dalkeith. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre's responsibility to obtain the necessary copyright clearance. SQA's Assessment Materials Team at Dalkeith may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

2 General Marking Principles These principles describe the approach taken when marking Advanced Higher Mathematics papers. For more detailed guidance please refer to the detailed Marking Instructions. The main principle is to give credit for the skills demonstrated and the criteria met. Failure to have a correct method may not preclude a candidate gaining credit for their solution. The answer to one part of a question, even if incorrect, is accepted as a basis for subsequent dependent parts of the question. 3 The following are not penalised: working subsequent to a correct answer (unless it provides firm evidence that the requirements of the question have not been met) legitimate variation in numerical values / algebraic expressions. 4 Full credit will only be given where the solution contains appropriate working. Where the correct answer might be obtained by inspection or mentally, credit may be given. 5 Sometimes the method to be used in a particular question is explicitly stated; no credit will be given where a candidate obtains the correct answer by an alternative method. 6 Where the method to be used in a particular question is not explicitly stated, full credit will be given for an alternative valid method. In the detailed Marking Instructions which follow, marks are shown alongside the line for which they are awarded. There are two codes used, M and E. M indicates a method mark, so in question, M,, means a method mark for the product rule (and then a mark for each of the terms). E is shorthand for error. So for example, E, means that a correct answer is awarded marks but that mark is deducted for each error. page

3 Advanced Higher Applied Mathematics 006 Section A Mechanics A. r (t) = ( 3 t3 4t ) i (t ) j v (t) = (t t) i 4tj a (t) = (t ) i 4j If a is in the jdirection then t = 4 so v (4) = 6i 6j v (4) = = 6 A. When v = v max, x = 0, so Using v max = ωa = 4 ω v = ω (a x ) with a = 4 But v = v max = ω so gives v = ω ( 6 x ). ω 64 = ω ( 6 x ) x = 6 64 = 3 64 x = ± 3 3 i.e. the distance from O is metres. A3. (a) a L = gj, v L (0) = 0r, L (0) = 0 v L (t) = gt j Also r L (t) = gt 6 j a B = gj, v B (0) = 0, r B (0) = j v B (t) = gtj r B (t) = ( gt ) j Br L = gt 6 + gt j = 9gt + j 6 page 3

4 (b) When, so Br L = 0 gt = 3 9 r B = ( 3 9 ) j = 9 j Distance light bulb falls = 9 = 6 9 metres. A4. (a) Using x = 0, y = g, v = V cos αi + V sin αj x = V cos α t y = V sin α t gt, When y = 0, t = V g sin α (t > 0) Range, R = V cos α V g sin α = V g ( sin α cos α) = V sin α. g (b) With α = 5 and R > Land R < L, R > L R < L V V sin30 > gl gl > V gl > V gl < V i.e. < V gl <. gl < 4 V gl < A5. (a) u 3m 0 m v 4m P before = 3mu; P after = 4mv By conservation of momentum v = 3 4u Using v = u + at, 0 = 3 4u + at a = 3u 4T. and R = 4m 4T 3u R = 3mu T (b) Using s = ut + at gives Distance to rest = 3 4 ut 3u 4T T Work done = 3uT page 4 = 3 4 ut ( ) = 3uT 3mu T = 9 mu Nm

5 A6. T mg T = λx = mgx l l (a) Resolving forces vertically λx cos 45 = mg l mgx l = mg x = l 4 = l (b) The length of the string is x + l. Resolving horizontally m (x + l) cos 45 ω = T sin 45 m ( + 4 ) lω = mg l 4 + lω ω = l 4 = g g ( + 4 ) l page 5

6 A7. (a) R T q q W Resolving perpendicular to the plane W cos θ = R + T sin θ () Resolving parallel to the plane W sin θ + µr = T cos θ () From () R = W cos θ T sin θ. Substituting into () W sin θ + µ (W cos θ T sin θ) = T cos θ T (cos θ + µ sin θ) = (sin θ + µ cos θ) W T = T = sin θ + µ cos θ cos θ + µ sin θ W sin θ cos θ + µ + µ sin θ cos θ T = tan θ + µ + µ tan θ W. W (b) We require T < W i.e.( tan θ + µ + µ tan θ) W < W tan θ + µ < + µ tan θ tan θ ( µ) < µ tan θ < since 0 < µ < and 0 < θ < π Thus 0 < θ < π 4. page 6

7 A. θ L v u = 7 gl (a) (i) By energy conservation mu = mv + mgl ( cos θ) When θ = 45 mg.7 gl = v + gl ( cos θ) so v = ( 3 ) + cos θ gl () v = (3 + ) gl (3 + ) gl v = (ii) Resolving forces along the line of the string With θ = 45 T = mv L + mg cos θ T = m (3 + ) gl L + mg = mg {3 + + } = 3 ( + ) mg (b) When the string goes slack, T = 0 so v = gl cos θ, so, in () gl cos θ = ( 3 + cos θ ) gl 3 cos θ = 3 cos θ = θ = 0 page 7

8 A9. + x A h (a) (b) ma = mg mkv v dv dx = g kv v dv g kv = dx Let w = g kv dw = kv dv k dw w = x + c k ln g kv = x + c As v = 0 when x = 0, c = k ln g x = k ln g k ln g kv = k ln g g kv g kx = ln g kv e kx g = g kv ge kx kv e kx = g kv e kx = g (e kx ) v = g k ( ) e kx (c) When x = h final KE = mv = mg k ( e kh ) = mg k ( e 4 ) Work done against resistance as fraction of initial PE mg mgh k ( e 4 ) = mgh = 4 ( e 4 ) page

9 Section B Mathematics for Applied Mathematics B So A = ( Other valid methods of obtaining A( x ) ( y = z so x = 0, y =, z =. ) A ). 0 will be accepted. x + y = x + 3y + z = x + y + z = ( ) ( x ) ( y = A = 0 z 0 )( ) ( = 0 ), M, E M, B. y = ln ( + sin x) dy dx = so d y dx cos x + sin x M, ( + sin x)( sin x) cos x cos x = M, ( + sin x) = sin x ( + sin x) = ( + sin x). B3. S n = 6n (n + )(n + ) S n + = 6 (n + )(n + )(4n + 3) (n) = 4 ( + + +n ) = 3n (n + )(n + ) page 9

10 B4. cos y dy dx = y dy y = sec x dx M so ln y = tan x + c., When y =, x = 0 giving c = ln. Hence ln y ln = tan x, i.e. ln y = tan x y = e tan x. B5. + x = u x dx = du so x 3 + x dx = (u ) u du = (u/ u / ) du = 3 u3/ u / + c = 3 ( + x ) 3/ ( + x ) / + c = 3 (x ) + x + c B6. (a) 0 x e x dx = [x e x dx e x dx] 0 M, = [ x e x 4 e x ] 0 = e 4 e + 4 = 4 (e + ) (b) 0 x e x dx = [x e x dx] 0 0 x. e x dx = [ x e x ] 0 0 x e x dx = [ e 0] 4 (e + ) = 4 (e ) (c) (3x + x) e x dx = 3 0 x e x dx + 0 x e x dx 0 = 3 4 (e ) + 4 (e + ) = 4 (5e ) [END OF MARKING INSTRUCTIONS] page 0