2013 Applied Mathematics Mechanics. Advanced Higher. Finalised Marking Instructions

Transcription

1 0 Applied Mathematics Mechanics Advanced Higher Finalised ing Instructions Scottish Qualifications Authority 0 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial basis. If it is to be used for any other purposes written permission must be obtained from SQA s NQ Assessment team. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of eamination 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 NQ Assessment team may be able to direct you to the secondary sources. These ing Instructions have been prepared by Eamination Teams for use by SQA Appointed ers when marking Eternal Course Assessments. This publication must not be reproduced for commercial or trade purposes.

2 Part One: General ing Principles for Applied Mathematics Statistics Advanced Higher This information is provided to help you understand the general principles you must apply when marking candidate responses to questions in this Paper. These principles must be read in conjunction with the specific ing Instructions for each question. (a) (b) s for each candidate response must always be assigned in line with these general marking principles and the specific ing Instructions for the relevant question. If a specific candidate response does not seem to be covered by either the principles or detailed ing Instructions, and you are uncertain how to assess it, you must seek guidance from your Team Leader/Principal Assessor. ing should always be positive ie, marks should be awarded for what is correct and not deducted for errors or omissions. GENERAL MARKING ADVICE: Applied Mathematics Statistics Advanced Higher The marking schemes are written to assist in determining the minimal acceptable answer rather than listing every possible correct and incorrect answer. The following notes are offered to support ers in making judgements on candidates evidence, and apply to marking both end of unit assessments and course assessments. These principles describe the approach taken when marking Advanced Higher Applied Mathematics papers. For more detailed guidance please refer to the detailed ing 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, can be accepted as a basis for subsequent dependent parts of the question. 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 epressions. 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 eplicitly 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 eplicitly stated in the question paper, full credit is available for an alternative valid method. (Some likely alternatives are included but these should not be assumed to be the only acceptable ones.) In the detailed ing Instructions which follow, marks are shown alongside the line for which they are awarded. There are two codes used M and E. The code M indicates a method mark, so in question B, M means a method mark for understanding integration by parts. The code E refers to error, so in question B6(b), up to marks can be awarded but mark is lost for each error. Page

3 Part Two: ing Instructions for each Question: Section A Question Solution Ma Additional Guidance A A particle is moving in a plane such that t seconds after the start of its motion, the velocity is given by (ti 5 t j) ms -. The particle is initially at the point ( 7 ) i j metres relative to a fied origin. Find the distance of the particle from O when t 5 s vdt t i t j c t 0 s ( i 7 j) c ( i 7 j) 5 s ( t ) i ( t 7) j t = s = (i + 8j) s = A A ball of mass 0.5kg is released from rest at a height of 0 metres above the ground. If the ball reaches.5 metres after its first bounce, calculate the size of the impulse eerted by the ground on the ball. Method : s 0 t u 0 v? a g v u as u 0 g u ms - s.5 t u v 0 a g v u as 0 v 5 v 7 g I = mv mu I = 05 (7 - (-) I = 05 Ns Method : Initial E 5g P = 0.5 metres On impact: mu 5g u 0g u ms Final 5g 5g - EP mv v 5g 7ms I = mv mu I = 05 (7 (-) I = 05 Ns Page - M Integration of velocity for displacement with correct integration Evaluate c and give vector for displacement Find vector when t and its magnitude M Motion under gravity to find velocity on impact Value of u M Motion under gravity to find velocity of rebound M Impulse momentum equation M Energy equation for initial PE and impact KE Value of u M Energy equation for final PE and rebound KE M Impulse momentum equation

4 Question Solution Ma Additional Guidance A A particle of mass kilograms moves under the action of its own weight and a constant force F ( i 5. j) where i and j are unit vectors in the horizontal and vertical directions respectively. 5 Initially the particle has velocity ( i j) ms - as it passes through a point A. The particle passes through B after seconds. Find the work done to move the particle from A to B. Method : F = ma 0 a a 5 g 8 s = ut + ½at s M Collective force correct M Method and calculation of acceleration Use of stuva and correct substitution to find displacement Work done = F s= Method 6 68 = = 680J M Method and calculation of work done Correct answer F ma 0 a a 5. g 8 a = i 8j v = ti 8tj + c t = 0 v = i j v = ( + t)i (8t + ) j Work done = F vdt = 0 0 t dt 8t ( ) ( ) M Collective force correct M Method and calculation of acceleration Integration to find epression for v M Method and calculation of work done Correct answer [ ] Page

5 Question Solution Ma Additional Guidance A A go-kart of mass 00 kilograms accelerates at ms - at the instant when its speed is 5ms - and the engine s power is at a maimum. Given that there is a total resistance to motion of 60N throughout the go-kart s motion, find the maimum speed which the go-kart can achieve. P P F = v 5 P Accelerating force = 60 5 P F = ma P = 800W M Correct formula and substitute to find accelerating force Calculation of Power Maimum speed: P a = V ma P 800 0ms Vma M Understanding of maimum speed Calculation of speed. Page 5

6 Question Solution Ma Additional Guidance A 5 A piano of mass 60 kilograms is resting on a rough plane 7 inclined at an angle θ to the horizontal, where tan. When a removal man applies a horizontal force of 850 newtons, the piano is just on the point of moving up the plane. Find the value of the coefficient of friction between the piano and the surface of the plane 6 When the removal man increases the horizontal force to 000 newtons, the piano begins to accelerate up the plane, along the line of greatest slope. How far does the piano travel in seconds? R850sin 60gcos 7 = 850( ) 60(9 8)( ) 7 8N cos R60g sin 7 R 850( ) 60(9 8)( ) 76 96N 5 5 R R 78 M Correct diagram including friction, horizontal force, normal reaction and weight and method of resolving in perpendicular directions Correct resolution perpendicular to the slope Correct resolution parallel to slope Correct value of μ 7 R = 000( ) 60(9 8)( ) 785 8N F ma : 000( ) 60(9 8)( ) (785 8) 60a a 086ms s? t u 0 a 086 s ut at : s (086)( ) 8 metres Equilibrium perpendicular to slope and F ma along the slope to find acceleration stuva substitution to find displacement Page 6

7 Question Solution Ma Additional Guidance A 6 A rough disc rotates in a horizontal plane with a constant angular velocity ω about a fied vertical ais through the centre O. A particle of mass m kilograms lies at a point P on the disc and is attached to the ais by a light elastic string OP of natural length a metres and modulus of elasticity mg. 5 The particle is at a distance of 5 a from the ais and the coefficient of friction between P and the disc is. Find the 0 range of values for ω such that the particle remains stationary on the disc. mg( a / ) mg T l a Slipping out: R mg T R mr mg mg 5a m 0 mg 5ma 0 g 5a Slipping in: T R mr mg mg 5a m 0 7g 5a M Hooke s Law M vertically equilibrium and horizontally combines forces of elastic string and friction Correct value for ω M Correct interpretation for slipping in Calculation of ω and final statement No slipping if: 7g g 5a 5a Page 7

8 Question Solution Ma Additional Guidance A 7 A light elastic string of natural length l metres hangs from a fied point O with a particle of mass m kilograms attached at its lower end. In equilibrium the string is etended by e metres. 6 The particle is then pulled down a further distance a metres where a < e and released. Show that the ensuing motion is simple harmonic and state the period of the motion. The maimum velocity of the particle during motion is ge. Find an epression for the amplitude of the motion in terms of e. e In equilibrium: Tension mg l ml mg T m ( e ) mg m l mgl e mg ( ) m e l g ( e e) e g [ ] e ml mgl e M Use of Hooke s Law in equilibrium position M In etension F=ma and substitution Proof of SHM i.e SHM where g e M Statement of Period Period e ml [ ] g M Statement for v ma and substitution v ma aw g ge a e g ge a ( ) e a e Calculation of final answer Page 8

9 Question Solution Ma Additional Guidance A 8 A smooth solid hemisphere of radius a metres is fied with its plane face on a horizontal table and its curved surface uppermost. A particle P of mass m kilograms is placed at the highest point on the hemisphere and given an initial 6 ag horizontal speed ms -. The particle moves along the curved surface of the hemisphere until it leaves the surface at Q Calculate the angle between the tangent at Q and the horizontal, and find an epression for the speed of the particle at Q. P R Q Θ mg ga 5ga Energy at Initial position: EP EK mga m At Q: Total energy: EP EK mga cos mv Conservation of energy: 5mga mga cos mv M initial total energy stated M Energy at Q and conservation of energy 5ga v ga cos (i) At Q consider forces acting towards O mv mg cos R a When body leaves surface R 0 mv mg cos a v ga cos (ii) In (i) 5 ga ga cos ga cos M Apply F=ma towards O M Interpretation of body leaving surface as R = 0 (stated) Algebraic manipulation to find θ cos = 6 5 = 6º v 5ga 6 Substitution in (ii) to find v Page 9

10 Question Solution Ma Additional Guidance A 9 A speedboat has to round three buoys P, Q and R as part of a race, starting at P and travelling anticlockwise.. The buoys are 00 metres from each other with R due North of Q and P lying to the west of the line QR. In still water, the speedboat travels at 0ms -. The water current is steady at 5ms - flowing from due West. Find the mean speed for one complete lap of the course. 9 PQ: P θ 0 0 Q V C 0 sin sin V cos8 V T PQ PQ PQ 00 8 secs V PQ M Interpretation of journey PQ /annotated diagram Calculation of true velocity PQ Time for PQ QR: 0 5 R Q V T QR QR secs V QR M Interpretation of QR/annotated diagram Calculation of true velocity QR and time for QR RP: 5 0 P R sin sin M Interpretation of RP /annotated diagram ( this is more demanding hence repeated method mark) V V T cos 8 RP RP RP sec s 55 Total Time for lap: = 5 secs 600 Mean Speed per lap: 9ms 5 Calculation of true velocity RP and time for RP Total Time Mean speed Page 0

11 Question Solution Ma Addition Guidance A 0 Two projectiles are launched simultaneously from points A and B, where B is due East of A and situated on the same horizontal plane through A. The projectile launched from point A is projected towards B with speed 90ms - at an angle of 0 to the horizontal. The projectile from point B is projected towards A with speed 50ms - at an angle θ to the horizontal. The two missiles collide in mid-air at a distance d metres horizontally from point A. Show that the height h at this point of collision is d(050 gd) h 50 0 Find the angle of projection θ at which the projectile at B is launched. The projectiles collide 5 seconds after launch. Calculate the distance between A and B. Projectile A: d 90cos0 5 t d d t 5 5 s h t t u 90sin0 5 a g d g d s ut at : h 5( ) ( ) 5 5 h d gd 50 d(050 gd) h 50 Projectile B : d(050 gd) d s t 50 5 u 50sin a g s ut at d(050 gd) d g d 50sin ( ) ( ) (050 gd) 0 d gd sin sin Horizontal displacements: d 5 t 5 [ 897] A 9 B 50 cos t [ 09 0] 0 Distance between A and B = m M Horizontal motion with constant speed to give epression for t M Vertical motion under gravity with values for stuva stated Epression for h Manipulation to give answer M Vertical motion under gravity with values for stuva stated Algebraic substitution and manipulation Epression for sin and value of θ M Epressions for horizontal distances Final answer Page

12 Question Solution Ma Addition Guidance A A body of fied mass m kilograms is projected vertically upwards from a point on the surface of a planet with an initial speed of u ms -. Assuming that the gravitational force GMm on the body is where d metres is the distance from the d centre of the planet, show that the speed of the body when it has reached a height h metres above the surface is given by v GMh u, where M kilograms is the mass of the R( R h) planet, R metres is the radius of the planet, and G is the gravitational constant. Find an epression for the maimum height H reached by the body. 0 Show that the escape speed necessary for the body to continue GM into space can be written in the form u k and state R the value of k. GMm GM dv F=ma: m acc v ( R h) ( R h) dh h = 0, v = u: GM dh vdv ( R h) GM v R h c GM u c c GM u R R GM v GM u R h R GM GM GM ( R ( R h)) v u u ( R h) R R( R h) v u GMh R( R h) Ma height: v 0; h H u GMH GMH 0 u R( R H ) R( R H ) u R( R H ) GMH u R u RH GMH u R H GM u R Ru H GM Ru Escape speed: GM u k R ( ) H GM Ru 0 M Use of F=ma and appropriate substitution M Method of separate variables Integration and substitution Epression for c using initial conditions Rearrangement of formula M Interpretation of ma ht by substituting v=0 Algebraic manipulation Correct answer M Understanding of escape speed with substitution Manipulation and value of k Page

13 Section B Question Sample Answer/Work Ma Criteria for dy B Given that y = sin (e 5 ), find d dy d = cos e 5 d ( e 5 ) d First application of chain rule. = cos e 5 5e 5 = 5e 5 cos e 5 Second application of chain rule. Notes: Page

14 Page Question Sample Answer/Work Ma Criteria for B Matrices are given as A = 0 B = 0 5 C = y B a Write A B as a single matri B A = A correct. = = 0 6 For correct evaluation of B and simplify. B b (i) Given that C is non-singular, find C -, the inverse of C. det C = y + Determinant correct. C - = y y Inverse correct. B b (ii) For what value of y would matri C be singular? y + = 0 for C singular y = y value correct. Notes:

15 Question Sample Answer/Work Ma Criteria for B Use integration by parts to obtain ln d where > 0 u ln, dv d du = M Understand integration d, v = d by parts. v I ln.. d Integrates dv and substitutes correctly. ln d Correctly combines v and = du. ln Correctly integrates = c second term. Notes:. Treat omission of c as bad form: do not penalise.. Negative indices for equally acceptable. Page 5

16 Question Sample Answer/Work Ma Criteria for B a State the results for r and r in terms of n. n r n r Hence show that n ( r r r) n (n)(n )(n ) n n( n ) r r n r r n ( n ) Both formulae correct. n ( r r) = r r Correct separation. r n r n r = n ( n) n( n) Substitution. n( n ) [ n( n ) 6] = n ( n)( n n6) Algebra correct. Note: This or equivalent intermediate algebra required for this mark. Notes: Page 6

17 Question Sample Answer/Work Ma Criteria for B (cont) B b Use the above result to evaluate (r 5 r 5 r) 5 ( r r) ( r r) ( r r 5 5 r r r) Correct limits. = = = 970 Correct evaluation. Notes: Page 7

18 Question Sample Answer/Work Ma Criteria for B 5 Find the general solution of the differential equation 6 dy y 6, 0 d dy y 6 d Multiplies through by. I.F = e = e ² Correct integrating factor. e ² d y + e ² y = 6. e ² d d (e ².y) = 6. e ² Recognises LHS as eact d differential of g I.F. d d (e ².y) d = 6. e ² d Knows to integrate. e ². y = e ² + c Integrates correctly. c y e Divides through by e ². Notes: 5. Final answer of y = + ce -² also correct. 5. c required for mark here. Page 8

19 Question Sample Answer/Work Ma Criteria for B 5 Alternative: d y 6 y d dy 6 y d Separates variables. ln 6 y k Integrates LHS. Integrates RHS (constant on either side). ln 6 y k Prepares for eponential. 6 y Ae Converts form to eponential., y Ae 6 y Ae Rearranges to make y subject.,5 C y e Notes: 5. Any constant acceptable. Therefore, term containing constant can be positive or negative y = e -²-c a valid alternative for this mark. 5.5 Either of last two lines valid for award of final mark. Page 9

20 Question Sample Answer/Work Ma Criteria for B 6 The cycloid curve below is defined by the parametric equations = t sin t, y = - cos t. y cycloid O B 6 a dy Find in terms of t d dy d = sin t, = cost t dt dt Appropriate differentiation. dy dy d sin t d dt dt cos t Correct use. Notes: Page 0

21 Question Sample Answer/Work Ma Criteria for B 6 (cont) B 6 b Show that the value of where 0 < t < d d y is always negative, in the case 5 d y d dy d d dt d dt M Correct application of method. = ( cos t)cos t sin t (sin t) ( cost) ( cos t) cos t cos t sin ( cos t) t E Differentiates /substitutes correctly. [(cos t sin t) cos t] ( cos t) Uses sin t + cos t = and simplifies. ( cos t) ( cos t) ( cos t) 0 Clear eplanation. Hence d y < 0, for 0 < t < d Notes: Page

22 Question Sample Answer/Work Ma Criteria for B 6 c A particle follows the path of the cycloid where t is the time elapsed since the particle s motion commenced. Calculate the speed of the particle when t = π. Speed = d dy dt dt Correct formula. Applies correct values to obtain a speed of. Notes: [END OF MARKING INSTRUCTIONS] Page