National Quali cations SPECIMEN ONLY

AH National Quali cations SPECIMEN ONLY SQ5/AH/0 Mathematics of Mechanics Date Not applicable Duration hours Total marks 00 Attempt ALL questions. You may use a calculator. Full credit will be given only to solutions which contain appropriate working. State the units for your answer where appropriate. Any rounded answer should be accurate to three significant figures (or one decimal place for angles in degrees) unless otherwise stated. Write your answers clearly in the answer booklet provided. In the answer booklet, you must clearly identify the question number you are attempting. Use blue or black ink. Before leaving the examination room you must give your answer booklet to the Invigilator; if you do not, you may lose all the marks for this paper. *SQAH0*

FORMULAE LIST Newton s inverse square law of gravitation GMm F = r Simple harmonic motion v = ω ( a x ) x= asin( ωt + α) Centre of mass Triangle: Semicircle: along median from vertex. r along the axis of symmetry from the diameter. π Standard derivatives Standard integrals f ( x) f ( x) f ( x) f ( x) dx tan x sec x sec ( ax) a tan( ax) + c cot x cosec secx secxtan x x x ax e ln x + c a e ax + c cosecx ln x cosecxcot x x e x e x Page two

Total marks 00 Attempt ALL questions MARKS Candidates should observe that g m s denotes the magnitude of the acceleration due to gravity. Where appropriate, take its magnitude to be 9 8 m s. 0. A curling stone, P, of mass 8 kg is moving with velocity m s relative to a suitable set of coordinate axes. It collides with a stationary curling stone, Q, of 06 mass 0 kg. Q then moves off with velocity 0 7 m s. Calculate the speed with which P travels immediately after impact.. Given y = e x cos x find dy dx.. A cyclist climbs a hill of length km and constant gradient :5 in minutes. When cycling at a constant speed of v m s, the external resistances to motion are (0 5 + 0 v ) N. Given that the total mass of the cyclist and bike is 66 kg, calculate the power produced by the cyclist.. A train travels from Glasgow to Stirling. It starts from rest and accelerates uniformly for the first 9 km of its journey. It then travels for 6 8 km at a uniform velocity, before decelerating uniformly to rest in 7 km. The total journey time is minutes. (a) Sketch a velocity time graph with appropriate units to represent this journey. (b) Calculate, in km h, the maximum speed reached by the train. (c) State one assumption you have made in answering this question. 5. A particle is projected vertically upwards at a speed of 8 m s from a point O. Two seconds later a second particle is projected vertically upwards from O at a speed of 5 m s. (a) Calculate the value of t when the two particles collide assuming that the only force acting is that due to gravity. (b) Determine the distance the particles are above O when they collide. Page three

6. An object moves horizontally along the x-axis with simple harmonic motion about a point O. The period of the oscillation is seconds. It is released from its extreme position A, a distance of metres from O. Find the first time the particle will be a distance of metres from A. MARKS 7. Calculate the gradient of the tangent to the curve xy xy = 5 at the point (,5). 8. A stone of mass 0 5 kg moves along a smooth horizontal surface. It passes a fixed point O with a velocity of i m s. When it is at a distance x metres from O, a force of magnitude ( sin x + 5)i N acts on it along the line of movement. (a) Calculate the work done by the force in moving the stone metres beyond O. (b) Find the speed of the stone at this point. 9. A house is being re-roofed. The old tiles slide down a rough plastic chute into a skip at the side of the house. The chute is 0 metres long and inclined at an angle of 0º to the horizontal as shown. 0 m 0º A tile of mass m kg is given an initial speed of m s at the top of the chute. The coefficient of friction between the tile and the chute is. Show that the speed of the tile at the bottom of the chute is 5 m s. 5 0. Find the exact value of using the substitution u = x +. 5 Page four

MARKS. A flower vase is modelled by rotating part of the curve π radians about the y-axis as shown in the diagram. x = 00 y through y O x (a) Find the volume of water needed to fill the vase to a depth of 6 cm. (b) State one improvement that could be made to the design. P Q 0 X O 6 R 6 S A uniform lamina is made from a rectangle OPQR and a right-angled triangle RQS. A circular hole of radius cm is removed as shown in the diagram. The centre X of the circular hole is cm from both OS and OP. (a) Taking O as the origin, find the coordinates of the centre of mass of the lamina. (b) When the lamina is suspended from a point T on OP, it hangs in equilibrium with OS vertical. Give the length of OT. 5 Page five

. The distance of the Earth from the Sun is 50 0 metres. The distance of Venus from the Sun is 08 0 metres. Calculate the period of rotation of Venus around the Sun, giving your answer in Earth years. State one assumption you have made when calculating your answer. MARKS 6. Three vessels A, B and C are being tracked by coastguards at half-hour intervals. With distances measured in kilometres and speeds in kilometres per hour, they have the following displacement and velocity vectors: Vessel A B C Time 0:00 0:0 :00 Position i + 7 j 6 i + 9 j i + 9 j Velocity i + 5 j i + j i + 6 j (a) Show that if A and C continue without changing course they will collide. Find the time and position of the collision. At the instant of the collision, vessel B changes course and then proceeds directly to the scene of the collision at its original speed. (b) Find the time, to the nearest minute, at which vessel B will arrive at the scene of the collision and state the bearing of its course to this point. 5 5 5. A golfer hits a ball from the point O with velocity (Pi + Qj) m s. The ball first hits the ground a distance of 50 metres from O in the horizontal plane. (a) Show that PQ = 5 g. (b) Given that the ball passes through 5i + 6j (i) Calculate P. (ii) Calculate the initial angle of projection to the horizontal. Page six

6. The movement of a door-closer on a hinged door is modelled by the differential d y dy equation 8 + 6y = 0. dt dt (a) Find the solution y = f (t) to this differential equation, given that y = and dy dt = when t = 0. (b) State which type of damping is described by the motion and give a reason for your answer. MARKS 6 7. A car of mass M kg is travelling with a speed of v m s round a circular bend of radius 0 metres on a road banked at 0º to the horizontal. The coefficient of friction between the car tyres and the road surface is μ. (a) Show that the square of the maximum speed the car can travel without slipping is given by 9( + μ) v = μ The minimum speed that the car can travel round the bend without slipping is u m s. (b) Given that v = u, calculate the coefficient of friction between the car tyres and the road. 5 6 [END OF SPECIMEN QUESTION PAPER] Page seven

AH National Quali cations SPECIMEN ONLY SQ5/AH/0 Mathematics of Mechanics Marking Instructions These Marking Instructions have been provided to show how SQA would mark this Specimen Question Paper. 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 purpose, written permission must be obtained from SQA s Marketing team on permissions@sqa.org.uk. Where the publication includes materials from sources other than SQA (ie 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 user s responsibility to obtain the necessary copyright clearance.

General Marking Principles for Advanced Higher Mathematics of Mechanics 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 Detailed Marking Instructions, which identify the key features required in candidate responses. (a) Marks for each candidate response must always be assigned in line with these General Marking Principles and the Detailed Marking Instructions for this assessment. (b) Marking should always be positive. This means that, for each candidate response, marks are accumulated for the demonstration of relevant skills, knowledge and understanding: they are not deducted from a maximum on the basis of errors or omissions. (c) Candidates may use any mathematically correct method to answer questions except in cases where a particular method is specified or excluded. (d) Working subsequent to an error must be followed through, with possible credit for the subsequent working, provided that the level of difficulty involved is approximately similar. Where, subsequent to an error, the working is easier, candidates lose the opportunity to gain credit. (e) Where transcription errors occur, candidates would normally lose the opportunity to gain a processing mark. (f) Scored-out or erased working which has not been replaced should be marked where still legible. However, if the scored-out or erased working has been replaced, only the work which has not been scored out should be judged. (g) Unless specifically mentioned in the Detailed Marking Instructions, do not penalise: working subsequent to a correct answer correct working in the wrong part of a question legitimate variations in solutions repeated errors within a question (h) Any rounded answer should be accurate to three significant figures (or one decimal place for angles given in degrees) unless otherwise stated. If an answer differs due to rounding or prior rounding the candidate may be penalised. Only penalise one mark in any question. Page two

Definitions of Mathematics-specific command words used in this Specimen Question Paper Determine: determine an answer from given facts, figures, or information. Find: obtain an answer showing relevant stages of working. Hence: use the previous answer to proceed. Hence, or otherwise: use the previous answer to proceed; however, another method may alternatively be used. Show that: use mathematics to show that a statement or result is correct (without the formality of proof) all steps, including the required conclusion, must be shown. Sketch: give a general idea of the required shape or relationship and annotate with all relevant points and features. Solve: obtain the answer(s) using algebraic and/or numerical and/or graphical methods. Page three

Detailed Marking Instructions for each question Question Expected response (Give one mark for each ) Max mark Ans: 0 5 m s - conservation of linear momentum calculate v calculate speed as the magnitude of velocity Additional guidance (Illustration of evidence for awarding a mark at each ) mu + mu= mv + mv 0 0 0 6 8 + 0 = 8v + 0 0 0 7 v 0 = 0 ( 0 ) + ( 0 ) = 0 5 m s - x Ans: e ( xcos x sin x) know and use chain rule know and start to use product rule complete process xe x x xe cos x +... x x xe cos x+ e ( sin x) Ans: 6 W calculate velocity of cyclist find force produced by cyclist to overcome gravity 00 0 v = = m s - 60 F = 66gsin θ +... find total force produced by cyclist 0 F = 66gsin θ + ( 0 5 + 0 ) = 0 8 use P = Fv to find power Note: and are still available to candidates who miss. 0 P= Fv: P= 0 8 = 6 W Page four

Question Expected response (Give one mark for each ) Max mark a Ans: Additional guidance (Illustration of evidence for awarding a mark at each ) 0 55 sketch velocity time graph v-t graph with 0 at the origin and 0 55 at the RHS of the t-axis (t hrs). correct annotations labelling of areas under the graph (9 km, 6 8 km and 7 km) b Ans: v = km h - set up two appropriate equations of motion elimination of one variable 5 algebraic manipulation to find v tv= 9 and ( t t) v= 6 8 or ( t t) v= 6 8 and ( 0 55 t ) v = 7 either tv= 8 and tv tv = 6 8 tv = 6 8 or tv tv = 6 8 and 055 v tv= 055 v tv= 6 5 either ( 0 55 t ) v = 7 0 55 v 6 8 = or tv = 9 tv = 8 0 55v 8 = 6 6 calculate maximum speed 6 v = km h - c Ans: one correct assumption 7 correct assumption 7 train is treated as a particle Page five

Question Expected response (Give one mark for each ) Max mark Additional guidance (Illustration of evidence for awarding a mark at each ) For accept (t mins) with labelled. For accept use of labels for unknown velocity and times at which train is travelling at maximum speed (for example, v, t and t ). 7 is not available if a candidate gives more than one assumption. 5 a Ans: t = consider motion of first particle under constant acceleration s= ut+ at s = 8t 9t consider motion of second particle under constant s = ut + at acceleration and substitute time s = 5 ( t ) 9( t ) relative to first particle equate displacements 8t 9t = 5 ( t ) 9( t ) 8 t 9 t = 5 t 0 9 t + 96 t 96 8t= 8t 0 calculate t t = 5 b Ans: s = 0 m (or equivalent) 5 calculate distance above O 5 s = 8 9 = 0 m 6 Ans: t = 65s Note: find the period of the function π ω =, T π ω = 6 correct formula and begin to π substitute correctly cos t =... 6 complete substitution and begin to solve use radians to find a value and find value for t cos t = cos π 6 π t = 6 π t = 6 9, t = 65s Accept correct solution arrived at using alternative formula x= asin( ωt+ α). Page six

Question Expected response (Give one mark for each ) Max mark Additional guidance (Illustration of evidence for awarding a mark at each ) 7 Ans: 5 6 start to differentiate dy dy y + xy +... or... y x dx dx complete differentiation dy dy y + xy y x dx dx = 0 expression for dy dx dy y y = dx xy x or equivalent value of gradient dy dx 5 5 5 = = 5 6 8 a Ans: 9 0 J b work done = f ( xdx ) with substitution a 0 ( sin x + 5 ) dx integration and answer [ cos x+ 5x] = 9 0J 8 b Ans: 7 m s - 0 work done = change in energy calculation of speed increase in EK = m( v u ) 9 0 = (0 5)( v ) v = 7ms Alternative solution using Newton s second law This solution gives the answer for (b) before (a) but is acceptable as long as the whole solution is in logical order and the answers for each part clearly identified. 8 b Ans: 7 m s - use of F = ma with appropriate substitution F = ma: sinx+ 5= 0 5v dv dx integration and completion 0 5vdv = ( sin x + 5) dx v = 5x cos x+ c 8 5 x= 0, v= : c= v = 0x 6cosx+ 5 8 x= : v = 0 6cos( ) + 5 v= 7 m s - Page seven

Question Expected response (Give one mark for each ) Max mark 8 a Ans: 9 0 J Note: For 8a accept 9 J. increase in KE = work done calculate work done Additional guidance (Illustration of evidence for awarding a mark at each ) Work = ( ) mv u = ( 05 )( 7 ) = 90J 9 Ans: proof 5 form equation using resultant mg sinθ μn = ma force find normal reaction force N = mgcosθ substitute all values into mg mg equation =ma find acceleration g a = 5 use STUVA equation and complete 5 g v = + 0, v = 5 and complete 9 Alternative find total energy at top 5 m + mg 0sin0= m+ 5 mg find total energy at bottom and hence the change in energy use frictional force to find work done use work energy principle loss in E = m+ 5mg mv k ( cos ) μ N d = mg 0 0 = 5 mg 5mg = m( v ) 5 complete proof 5 v = 5 and complete 9 Alternative use W = Fd with correct resultant force substitute into W = Fd and simplify expression for work done find expression for change in kinetic energy 5 = ( sinθ μ cosθ ) W mg mg d 5mg mg = 0 ( sin 0 cos0) mv m = m( v ) Page eight

Question Expected response (Give one mark for each ) use work energy principle Max mark 5mg Additional guidance (Illustration of evidence for awarding a mark at each ) = m( v ) 5 complete proof 5 v = 5 and complete 0 Ans: = 8 5 change variable and differentiate u = x + du = dx change the limits when x = 7, u = 9 and when x =, u = write the integral in terms of u 9 9 u du u u du u ready to be integrated / / = ( ) integrate u = u / 5 obtain correct answer 5 = 8 0 Alternative solution / 9 = 6 ( ) A candidate may choose not to change the limits, as illustrated in mark. In this case mark as follows: change variable and differentiate write the integral in terms of u ready to be integrated integrate substitute in equation for x 5 obtain correct answer 5 = 8 a Ans: 750 cm 5 u = x + du = dx u du = ( u u ) du u u = u ( x + ) = ( x + ) = 6 ( ) 7 state integral to be used to find volume interpret diagram to state limits V = πx dy = π( 00 y ) dy 6 0 π( 00 y ) dy Page nine

Question Expected response (Give one mark for each ) complete integration Max mark calculate integral 750 Additional guidance (Illustration of evidence for awarding a mark at each ) y 6 V = π[ 00 y ] 0 b Ans: valid improvement 5 state improvement 5 eg vase would need a flat base a Ans: C( 7, 50) 5 state area and position of CoM for one of the regular shapes state area and position of CoM for other two regular shapes Area From OP From OS OPQR 60 5 QRS 0 8 0 Circle π Lamina 90 π x y take moments about OP equating to moment of lamina 60 + 0 8 π = ( 90 π) x find x coordinate 0 π x = = 7 90 π 5 take moments about OS to find y coordinate 5 0 60 5 + 0 π = ( 90 π) y b Ans: OT = 50 6 interpret rotation 6 50 Ans: 0 6 years 6 state any valid assumption calculate angular velocity of Earth eg orbits are circular, any force created by the rotation of the planets can be ignored Earth year = 56000 seconds, hence π 7 ω E = 99 0 60 60 65 Page ten

Question Expected response (Give one mark for each ) use inverse square law and centripetal force for either planet use method for second planet Max mark Additional guidance (Illustration of evidence for awarding a mark at each ) GMm mr E E Eω E = r or and calculate constant GM re ωe ( 5 0 ) 5 calculate angular velocity of E GMm mr V Vω V = r V V π = = 56000 0 = 97 0 GM 97 0 = = Venus 5 V r V ( 08 0 ) 6 calculate period in Earth years ω so ωv = 6 0 7 6 π T = = 9666 s = 0 6 7 6 0 years Note: A correct answer using angular velocity in radians per earth year instead receives full credit. 0 a Ans: 00, 8 5 7 find position vector of A at 00 6 + = 7 5 express displacement for A and 6 6+ t C after t hours + t = 5 + 5t and + t + t = 9 6 9+ 6 t prove collision 6+ t = + t t = + 5t + 9 + 6t t = proving displacements same after hours state time at collision collision: 00 5 state displacement vector at collision 5 collision: position: 8 7 b 5, 00 6 5 6 state position of B at 00 7 find distance from B to collision point 8 find time to collision point using speed of B 6 6 7 6 5 + = 9 7 BK = ( 8 6 5) + ( 7 ) = 7 km 8 Time = 7 = 0 85 hrs = 5 mins 5 Time of arrival 5 Page eleven

Question Expected response (Give one mark for each ) 9 identify method to find bearing Max mark Additional guidance (Illustration of evidence for awarding a mark at each ) 9 7 tan ( ) = 69 8 6 5 0 calculation of bearing 0 Bearing 00 6 Alternative 6 state position of B at 00 6 6 7 6 5 + = 9 7 statement for displacement in 6 5 5sinθ 8 terms of bearing 7 + t = 5cosθ 7 8 solve equations for θ 5tsinθ = 5 5tcosθ = 8 5 tanθ = θ = 0 6 9 state bearing 9 bearing 00 6 0 calculation of time 0 t = = 5 mins 5cos0 6 5 a Ans: show PQ = 5 g equate the vertical component of the position to zero, or the vertical component of the velocity to zero determine an expression for the time of flight substitute into relevant equation rearrange to the required result time of arrival 5 Qt gt = 0 or Q gt=0 Q t = g Pt = 50 Q P = 50, PQ = 5 g g Alternative solution mark as above with only different. Pt = 50 and Qt gt = 0 50 50 50 t = Q g = 0 P P P 50Q 500g = 0 P P 00PQ 500g = 0 PQ = 5 g Page twelve

Question 5 b i Ans: Expected response (Give one mark for each ) P Max mark = 6 5 ms 5 set up equations for the horizontal and vertical components of the position 6 substitute equation for t into the equation for the vertical component of the velocity Additional guidance (Illustration of evidence for awarding a mark at each ) 5 Pt = 5 and Qt gt = 6 5 6 5 5 t = Q g = 6 P P P 90PQ 05g = P 7 substitute to get equation for u 7 PQ =5 g 90( 5g) 05g= P 50g 05g= P 8 solve to find value of u 8 P = 05, P = 6 5 m s - ii Ans: α = 9 6 9 evaluate the vertical component of the velocity 0 calculate the angle 9 PQ =5 g Q = 9 m s - P tanα = Q 9 0 tanα = 6 5 α = 9 6 6 a Ans: t t y= e te 6 correct auxiliary equation m 8m+ 6 = 0 correct solution of auxiliary equation ( m ) = 0 m = statement of general solution/complementary function y Ae t = + Bte t correct evaluation of A = A. + 0 A = 5 correct differentiation of general solution 6 substitution to obtain B and particular solution dy Ae Be Bte dt = + + 5 t t t 6 = Ae + Be + B. 0. e =. + B B = 0 0 0 t t y = e te Page thirteen

Question Expected response (Give one mark for each ) Max mark 6 b Ans: correct statement 7 type of damping Additional guidance (Illustration of evidence for awarding a mark at each ) 7 critically damped 8 correct explanation 8 since there is a repeated root 7 a Ans: show that 9( + μ) v = μ identify frictional force acts down slope resolve forces horizontally 5 frictional force acts down the slope (accept diagram) mv Rsin 0 + μrcos0 = r resolve forces vertically Rcos0 μrsin 0 = mg know to solve simultaneously 5 rearrange to required answer 7 b Ans: μ = 0 0 6 + μ v = μ 9 5 9( + μ) v = μ 6 resolve forces given that friction is now acting up the slope 6 mu Rsin0 μrcos0 = r Rcos0 + μrsin 0 = mg 7 obtain an equation for u 7 9( μ) u = + μ 8 equate v andu given v= u 8 58( μ) 9( + μ) = + μ μ 9 obtain a quadratic 9 8 μ 0μ+ 8 = 0 0 use the quadratic formula to obtain two solutions state required answer with reason 0 ± 8 0 μ = 6 μ = 0 0 or μ = 8 μ = 0 0 since 0 µ. can be implied by and. For candidates who resolve parallel and perpendicular to the slope only and are available. [END OF SPECIMEN MARKING INSTRUCTIONS] Page fourteen