2014 Applied Mathematics - Mechanics. Advanced Higher. Finalised Marking Instructions

Transcription

1 04 Applied Mathematics - Mechanics Advanced Higher Finalised ing Instructions Scottish Qualifications Authority 04 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 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 NQ Assessment team may be able to direct you to the secondary sources. These ing Instructions have been prepared by Examination Teams for use by SQA Appointed ers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

2 Part One: General ing Principles for Applied Mathematics Mechanics 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. 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 Mechanics 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. 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 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.) Page

3 Part Two: ing Instructions for each Question Section A Question Expected Answer(s) Max A I Ft mv mu 3(4) v 0 v 6ms - Conservation of Linear Momentum: m u m u ( m m ) v (6) m(0) ( m)(375) m kg 375 ALTERNATIVE SOLUTION F ma 3 a 3 a ms - 4 M: Use of impulse to calculate velocity of impact E: Value for velocity of impact M: Conservation of Linear Momentum E: Value of m M: Use of Newton's nd Law to calculate acceleration u = 0 t = 4 a = 3 v = u + at = (4) = 6 m s- E: Stuva to calculate velocity of before impact Conservation of Linear Momentum: m u m u ( m m ) v (6) m(0) ( m)(375) m kg 375 M: Conservation of Linear Momentum E: Value of Page 3

4 A 4 5 T 5 7 v ( a x ) ( a ) 5 49 a 5 a 37 metres x Asint 5t 37sin 7 t 046 seconds 4 E:Value of M: Correct formula for velocity and amplitude and correct substitution E: Value for amplitude M: Use of formula to find displacement and answer Page 4

5 A 3 (i) (ii) 00 W ( F 500) dx 0 00 (3000 5x 500) dx 0 5x 500x 00 0 M: Method of finding work done (with limits or calculation of constant included later) J = 00 kj E: Correct answer Work- Energy Principle: W mv mu v v 39m s M: Use of Work Energy Principle and substitution E: Correct calculation of speed Alternative solution for (ii) F F ma a m dv (500 5 x) dx v dx 700 F ma dv v 500 5x dx 700 vdv (500 5 x) dx 700 M: Use of correct differential equation and substitution v 5x v 39 ms 00 0 E: Correct calculation of speed Page 5

6 A 4 Equilibrium T cos30 T cos50 g T cos 30 T cos 50 g g T 85N cos30 cos50 T 65N mv F r v T sin 30 T sin 50 r r tan50 r v T sin 30 T sin v 0358( ) - v 6 m s 6 M: Consider equilibrium involving both tensions and weight E: Correct substitution of components E: Using conditions to find tension M: Horizontal use of mv F r (Consistent with M above) E: Calculation of radius of circle E: Algebraic manipulation to find v Note: If angular speed used can achieve 3/4. Page 6

7 A 5 Perpendicular to slope: Along slope: F = ma 4Mg R Mg cos 5 6 M: Equilibrium perpendicular to slope with equation R Mg sin Ma 4Mg 3Mg Ma g a ( 784) 5 Motion under constant acceleration up slope to rest: v u as 8gs 0 u 5 5u s 8g Consider motion down slope: F ma Mg cos Mg sin Ma 4 g a ( 39) 5 Constant acceleration down slope: v u as 4gs 4u 5 5u s g Distance AC = 5 u 5 35 u u g 8g 8g M: F ma along slope with equation E: Calculation of acceleration E: Calculation of displacement up slope to rest E: Calculation of acceleration down slope E: Calculation of displacement down slope and distance AC Page 7

8 A 6 Method : 6 r v P P 0 5cos 45 rs 0 5sin 45 0 cos 0 vs 0sin 0 M: Statements of displacements and velocity vectors at 3pm After time t 0 0 cos 0t cos rp + t = 0 0sin 0tsin 5cos r s + t 5sin t E: Statements of displacements after t hours At interception rp r s 0tcos 06 and 0tsin 06 0t t t 0 cos 0sin 0 0sin 0 0 cos sin cos sin( 45) 657 patrol vessel should steer on bearing 043 M: Equate components E: Algebraic manipulation E: Interpret answer to state bearing of interception 06 t 8 hours = hour 7 min 0cos65 7 Interception occurs at 4:7pm E: Calculation of time Page 8

9 A 6 (cont) Method : must be in the direction PS for interception θ M: For interception, relative velocity vector in direction PS V s =0 35 V p =0 M: Correct diagram annotated 0 0 sin35 sin sin 0sin35 = Patrol vessel must sail ( ) = 4 3º v = cos4 3 v = 6 5 t = hour 7 mins 6 Interception occurs at 4:7pm E: Use of trig E: Interpret answer to state direction of interception E: Find relative velocity E: Calculation of time Page 9

10 A 6 (cont) Method 3: 0 cos 0 vp vs 0sin 0 0cos pvs vp vs 0sin 0 v must be in the direction PS for interception p s 0cos cos 45 k 0sin 0 sin 45 k cos 45 0cos k sin 45 0sin 0 0sin 0 0 cos 0sin 0 0cos sin cos sin( 45) 657 patrol vessel should steer on bearing 043 M: Statement of condition for interception E: Expression for relative velocity vector M: Equate components E: Algebraic manipulation E: Interpret answer to state bearing of interception 06 t 8 hours = hour 7 min 0cos65 7 Interception occurs at 4:7pm E: Calculation of time Page 0

11 A 7 al g ab g 9 8g BaL g g M: Find relative acceleration 8g B vl B aldt t c 9 8g t 0, v 35 v t g B rl B vldt t 35 t k 9 4g t r r t t 9 0, B L 3 5 M: Use of calculus to find relative velocity E: Correct expression for relative velocity E: Correct expression for relative displacement Br ( ) 0 L t when ball hits floor 4g t 35 t 0 9 M: Statement for conditions when ball hits floor of lift 4gt 35t t 35t 9 0 E: Process of calculating time t 35 ( 35) E: Correct answer for time t 03 or t 0 t 03seconds (reject negative answer) Page

12 A 7 (cont) Second solution using relative acceleration and stuva al g ab g 9 8g BaL g g 9 9 8g s t u 35 v a 9 4g s ut at 35t t 9 4gt 35t t 35t 9 0 t 35 ( 35) t 03 or t 0 t 03 seconds (reject negative answer) Alternative solution Ball a g v gt c t 0 v 35 c 35 gt r 35 c t 0 r 0 c 0 gt r r r gt gt 8 35t 8gt 63t 8 0 4gt 35t t 35t 9 0 t ( 35) Lift g a 9 gt v k 9 t 0 v 0 k 0 gt r k 8 t 0 r k gt r 8 t 03 or t 0 t 03 seconds (reject negative answer) M Show understanding of motion under constant relative acceleration M: Find relative acceleration M :Consider motion under constant relative acceleration E: stuva and substitution E: Correct quadratic equation E: Process of calculating time E: Correct answer for time M Vertical motion of ball and lift separately E: Correct expression for displacement of lift/ball after t secs E : Correct expression for displacement of otherafter t secs M: Equating displacements E: Correct quadratic equation E: Process of calculating time E: Correct answer for time Page

13 A 8 (a) 3 (b) 7 At Q: total energy = mv 5 u At top of circle total energy: mgh mv 3 3 3g 8 v 54g v For complete circles v 0 : 5u 54 u Height at any time: 09( cos ) At rest (maximum height): Energy = mgh 3g 09( cos ) If u = 4 : Energy at Q = 4 3g 09( cos ) cos Angle of oscillation = 69 4 Maximum tension when mv T 3g r 3 4 T 3g 87N g 8g ms 5 - M: Consideration of conservation of energy. E: Correct statements of energy at bottom and top of circle M: For complete circles v 0 and find u M: General expression for height at any time M: Energy when rod is at rest E: Equate this with energy vertically below P E: Solve trig equation to find angle of oscillation M: Understanding of maximum tension (stated or implied) M: Use of mv F r E: Calculation of Tension Page 3

14 A 9 (a) (i) (ii) 3 t 3 t v adt 3 dt = 3 t c t 0,0 0 c 0 3 t v3 t t : v3 965 ms 8 3 : R 965cos5 t : s ut at 0 965sin 5t gt t( t) 0 t 0 or t 083 : R 965cos metres M: Integration to find expression for velocity E: Substitution and correct expression E: Substitute for t and correct answer for speed M:Consider motion horizontally and vertically with substitution E: Value of t E: Value of R A 9 (b) (i) (ii) cos t t 0cos : s ut t 0 sin 75 g cos 0cos 75tan sec 0 75tan tan tan 4 or tan : v u as s 45 or s 0 76m Athlete cannot jump 4 5m vertically Take-off angle 3 Page 4 3 M:Consider motion and expression for t M: Consider motion vertically with this value of t and substitution E: Solution of trig equation to give angles of projection E: Find two possible heights E: Explanation of answer

15 A 9 (b) (cont) Method : cos t t 0cos : s ut t 0 sin 75 g cos 0cos 75sin cos cos 75sincos sin sin [ cos ] = M:Consider motion and expression for t M: Consider motion vertically with this value of t and substitution E: Solution of trig equation to give angles of projection : v u as s 45 or s 0 76m Athlete cannot jump 4 5m vertically Take-off angle 5 E: Find two possible heights E: Explanation of answer Method 3 (equation of a trajectory): gx y xtan u cos g(75 ) 0 75tan (0 )cos 75tan sec 0 75tan tan tan 4 or tan : v u as s 45 or s 0 76m Athlete cannot jump 4 5m vertically Take-off angle 3 M:Consider equation of trajectory. E: When y 0 x 75 and u 0 arrange in suitable form and prepare to solve E: Solution of trig equation to give angles of projection E: Find two possible heights E: Explanation of answer Page 5

16 A 0 F kmg v kmg dv mg mv v dx 8 M: F P Connect v power and force and substitution. dv dx v g k v v dv k v ( ) v gdx kv k ln k v gx C E: F ma and using a v dv dx M: Integration to find displacement and separation of variables x 0 v 0: (0 ) k(0) k ln k 0 0 C E: Process of Integration C k ln k k gx k ln kv v k v E: Substitution and simplification At height h: k v u gh k ln ku u ku E: Final substitution and expression processed k mu mu k u mgh mu mk ln mku m k k ln k u ku k m k ln ku k u k k u ln mkg g k u g M: Work done = Change of energy E: Algebraic manipulation M: Work Time = Power E: Final manipulation Page 6

17 A 0 (cont) Alternative for last marks: Work done = T T T kmg Fvdt vdt kmgdt kmgt v k mu m( k ln ku u ) k u kmgt mu mgh k k u T ln g k u g [END OF SECTION A] Page 7

18 Section B (Mathematics for Applied Mathematics) Question Expected Answer(s) Max B y x x dy d d x. x x x dx dx dx x. x x Gradient given by dy when x = 0, dx Gradient = product rule first correct term second correct term evaluation (accept decimal equivalent to minimum of 3 sf) B (a) A B k evaluation B (b) 0 k k 0 det A = (0 + 3) 3(0 + 5) + 4(3k 0) = k form of determinant evaluation B (c) BC evaluation B (d) BC = 3I. B = 3C or C = 3B identity matrix connection or mention of inverse relationship correct Page 8

19 B 3 I xsin 3x dx 5 u = x du = dv = sin3x v sin3xdx cos3 x 3 evidence of integration by parts correct choice of u, dv I x. cos3x. cos3 3 xdx 3 correct substitution x cos3x cos3 3 3 xdx x cos3x sin 3x 3 9 final integration correct x I0 cos3x sin 3x 3 9 cos6 sin 6 0 sin evaluation Page 9

20 B r 3 r r r using n r n n n r * r 3 r = 5,640 correct substitution into * evaluation (using incorrect formula this mark available if of equivalent difficulty eg n r r nn B 5 (a) (e x + ) 4 x 4 x 3 x e e e x 3 x e. e x 3x x x e e e e Accept Binomial expansion or Pascal s Triangle correct coefficients correct powers of e x and simplification B 5 (b) x e 4 dx 4x 3x x x e 8e 4e 3e 6 dx 4x 3x x e 8e 4e x 3e 6x c 4 3 correct integration of composite function (at least one correct term involving composite exponential) completion of integral (+ c not essential) Page 0

21 B 6 (a) people. B 6 (b) 0000 A B N 0000 N N 0000 N = A(0 000 N) + BN A, B 5 appropriate form of partial fractions correct values of A and B Using N N dn dt gives dn dt N 0000 N separate variables Integrating, dn N 0000 N dt nn n 0000 N t c N n t c 0000 N starts integration eg dn correct N completes integration (moduli signs not required) Page

22 B 6 (c) N Using n t c 0000 N 4 N gives 0000 N e tc N 0000 N t c Ke where K e accurately converts to exponential form (stating explicitly K= e c not required) When t = 0, N = K K 99 Hence N 0000 N 99N = (0 000 N) e t N(99 + e t ) = 0 000e t t e 99 interprets initial condition K valve correctly gathers N terms 0000e N 99 e t t [END OF SECTION B] [END OF QUESTION PAPER] Page