2009 Applied Mathematics. Advanced Higher Mechanics. Finalised Marking Instructions

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1 2009 Applied Mathematics Advanced Higher Mechanics Finalised Marking Instructions Scottish Qualifications Authority 2009 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 the Question Paper Operations 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 Question Paper Operations 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 Advanced Higher Applied Mathematics 2009 Mechanics Solutions A. 0.m A O B (a) The maximum speed = 0 = ωa M ω = 0 0 = 00 Period of oscillation = T = 2π 00 = π 50 s (b) Now v 2 = ω 2 (a 2 x 2 ) M = 00 2 ( ) = = 75 When the particle is 5 cm from O, it has speed 8 66 m s. A2. Let the initial speed of the ball be V m s. The horizontal motion is given by x = (V cos θ) t 40 = 4Vt 5 The vertical motion is given by t = 50 V. Eliminating t gives y = (V sin θ) t 2 gt2 2 = 3Vt 5 2 gt2. 2 = g 2V 2 = 2500g 56 V 2 m s V 2 A3. Firstly, v p = 2ti + 4j and also v Q = 2ti 2 cos 2πtj + c M, Given that when t = 0, v Q = 0then c = 2j, so v Q = 2ti + 2 ( cos 2πt) j When the boats have the same velocity 2 ( cos 2πt) = 4 cos 2πt = t = 0 5, 5 s

3 A4. (a) dv Starting with = a and integrating gives v = at + c M When t = 0, v = u hence c = u and so v = u + at (*). ds Thus we have = u + at. Integrating gives s = ut + 2at 2 + k Since s = 0 when t = 0 then k = 0 so s = ut + 2at 2 (**) (b) From equation (*) t = v u a u (v u) (v u)2 and substitute into (**) to give s = + a 2a 2as = 2uv 2u 2 + v 2 2uv + u 2 = v 2 u 2 2E v 2 = u 2 + 2as A5. Method - velocities A km h 00 B - 70km h P sin PAB sin 00 By the sine rule = Hence PAB = 4. So the required bearing is 4 Method 2 - displacements A km 350t km 00 B 70t km P Let the time be t hours. Then sin PAB sin 00 sin 00 = sin PAB = 70t 350t 5 PAB = 4 So the required bearing is 4

4 A6. C Let the origin of a rectangular coordinate system be the point B and let i be the unit vector in the direction BA. Then B i A v A B = Ui and v B C = U cos 60 i + U sin 60 j = 2U (i + 3j) The impulse is given by I = m (v B C v A B ) I = 2mU (3i + 3j) and the magnitude of the impulse is I = 3mU A7. Let m kg be the mass of the block, and a m s its acceleration down the slope. h Then, ma = mg sin θ a = g sin θ. M Let s be the distance travelled down the slope. Now, sin θ = h, s s = h sin θ and also s = ut + 2at 2 = 2(g sin θ) t 2. h Hence. sin θ = g sin θ t 2 2 t 2 2h = g sin 2 θ. t = 2h g sin 2 θ.

5 A8. R µr mg Let R be the normal reaction force and µ the coefficient of friction between the cycle wheels and the track. Resolving forces in the vertical direction R cos 45 = µr sin 45 +mg R( µ) = 2mg ( ) Resolving in the horizontal direction mv 2 r = R sin 45 +µr cos 45 M R( + µ) = 3 2mg ( ) Dividing equations (*) and (**) + µ µ = 3 M µ = 2

6 A9. (a) A 2 L B L C Let T be the tension in AV and T 2 the tension in BC. Resolving horizontally gives T cos 60 = T 2 cos 30 T = 3T 2 ( ) Resolving vertically gives T sin 60 +T 2 sin 30 = W 3T + T 2 = 2W ( ) Using (*) and (**) to eliminate T we get 4T 2 = 2W T 2 = 0 5W. So the tension in BC is 0 5W newtons. M (b) Since CB = 3L, the extension of the string is ( 3 ) L. Using Hooke s law T 2 = λx L gives 0 5W = λ ( 3 ) W and the modulus of elasticity is. 2 ( 3 ) The elastic potential energy is E = λx2, 2L so E = 2L = 4 M W 2 ( 3 ) ( 3 )2 L 2 ( 3 ) LW joules

7 A0. A O 2 m q D R 5m 30 mg B C (a) Method - work-energy Height of Aabove BC is 5 sin 30 = 2 5 m. Change in potential energy = mg h = = 4 9 J M, Work done overcoming friction = Fs = = 0 4 J At C, kinetic energy of sledge = ( ) = 4 5 J Method 2 - Newton's laws Let abe the acceleration. Resolving parallel to AB ma = mg cos a = = 4 5 v 2 = = 45 2 mv2 = = 4 5 At C, kinetic energy of sledge = 4 5 J (b) Let D be the point at which the sledge leaves the track so that, as the sledge approaches D, the normal force, R, tends to 0. Let the speed of the sledge at Dbe v so that, at D, mg sin θ = mv2 v 2 = rg sin θ = 2g sin θ M r Hence at D, the kinetic energy of the sledge is mg sin θ = 96 sin θ. Height of Dabove BC = 2 sin θ + 2 cos 30 = 2 sin θ Potential energy at Dis mg(2 sin θ + 3) = 96 (2 sin θ + 732) By the conservation of energy, 96 (2 sin θ + 732) + 96 sin θ = 4 5 M 3 sin θ = sin θ = = Hence, the angle between OD and the horizontal is 0 8. M

8 A. (a) Whilst being towed, the equation of motion of the skier is 60 dv = 300 5v = 4 dv 20 v M t = 4ln 20 v + C When t = 0, v = 0, so C = 4ln20 and hence t = 4ln v 20 Rearranging gives 20 v = e0 25t 20 v = 20e 0 25t v = 20 ( e 0 25t ) When t = 6, v = 20 ( e 5 ) = 5 5 m s. The line BC passes through (6,5 5) and (0, 0). So an equation for BC is v 0 = 3 9 (t 0) v = t (b) Distance travelled between t = 0 and t = 0 is given by s = ( e 0 25t ) (39 3.9t) M = 20 [t + 4e 0 25t ] [39t.95t 2 ] 0 6 2E = 20(6 + 4e 5 4) + ( ) m [END OF MECHANICS SOLUTIONS]

9 Advanced Higher Applied Mathematics 2009 Section B Solutions B. ( b 2 b) 5 = b 5 + 5b 4 ( 2 b) + 0b3 4 b 2 ( + 0b2 8 b ) + 5b6 3 b 32 4 b 5 powers coeffs signs = b 5 0b b 80 b + 80 b 3 32 b 5 B2. u = cos x du = sin xdx, x = 0 u = ; x = π 3 u = 2 Hence OR π/3 0 cos 5 x sin xdx = 2 2 u5 du = 6 u6 = = 2 ( 0.64) 28 π/3 cos 5 x sin xdx = 0 6 cos6 x π/3 0 = = 2 28 B3. x = t 2 + dx y = 3t 3 dy dy dy dx = dx = 9t2 2t = 9t 2 = 2t 3E ( 0.64) = 9t 2 = 9 when t = 2. Point of contact is x = 5, y = 23. Equation of tangent is (y + 23) = 9 (x 5) y + 23 = 9x + 45 y + 9x = 22 M

10 B4. ( 0 det 0 k 2 2 k ) = k 2 2 k 0 k det ( ) det ( ) + 0 M, = (k 2) k + 2 (0 + ) = k 2 2k + = (k ) 2 = 0. Hence the matrix does not have an inverse when k =. B5. t dx 2x = 3t 2 dx 2 t x = 3t Integrating factor: 2 = 2lnt = ln t 2 so IF = t 2. t M, dx t 2 2 t 3x = 3 t x t 2 = 3 t = 3lnt + c x = t 2 (3lnt + c) (,) c = + 0 x = t 2 ( + 3lnt) B6. f (x) = x tan 2x π/ x tan 2x cos 2 2x f (x) = tan 2x + 2x sec 2 2x M, f (x) =2 sec 2 2x + 2 sec 2 2x + 2x(4 sec2x(sec2x tan2x)) 2E = 4 sec 2 2x + 8x sec 2 2x tan 2x = 4 sec 2 2x ( + 2x tan 2x). dx = 4 sec 2 2x ( + 2x tan 2x) dx, 4 π/6 0 = 4 [tan 2x + 2x sec2 2x] π/6 0 = π = 4 + π 3. [END OF SECTION B SOLUTIONS]