L.41/42. Pre-Leaving Certificate Examination, Applied Mathematics. Marking Scheme. Ordinary Pg. 4. Higher Pg. 21.

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1 L.4/4 Pre-Leaving Certificate Examination, 0 Applied Mathematics Marking Scheme Ordinary Pg. 4 Higher Pg. Page of 44

2 exams Pre-Leaving Certificate Examination, 0 Applied Mathematics Ordinary & Higher Level Table of Contents Ordinary Level Higher Level Q Q.... Q Q Q Q Q Q Q.... Q.... Q Q Q Q Q Q Q Q Q L.4/4_MS / Page of 48 exams

3 exams Pre-Leaving Certificate Examination, 0 Applied Mathematics Ordinary & Higher Level Explanation. Penalties of three types are applied to students work as follows: Slips - numerical slips S( ) Blunders - mathematical errors B( ) Misreading - if not serious M( ) Serious blunder or omission or misreading which oversimplifies: - award the attempt mark only. Attempt marks are awarded as follows: (Att. ), 0 (Att. ). Mark all answers, including excess answers and repeated answers whether cancelled or not, and award the marks for the best answers.. Mark scripts in red unless a student uses red. If a student uses red, mark the script in blue or black. 4. Number the grid on each script to 9 in numerical order, not the order of answering.. Scrutinise all pages of the answer book. 6. The marking scheme shows one correct solution to each question. In many cases, there are other equally valid methods. Current Marking Scheme Assumptions about these marking schemes on the basis of past SEC marking schemes should be avoided. While the underlying assessment principles remain the same, the exact details of the marking of a particular type of question may vary from a similar question asked by the SEC in previous years in accordance with the contribution of that question to the overall examination in the current year. In setting these marking schemes, we have strived to determine how best to ensure the fair and accurate assessment of students work and to ensure consistency in the standard of assessment from year to year. Therefore, aspects of the structure, detail and application of the marking schemes for these examinations are subject to change from past SEC marking schemes and from one year to the next without notice. 0. L.4/4_MS / Page of 48 exams

4 exams Pre-Leaving Certificate Examination, 0 Applied Mathematics Ordinary Level Marking Scheme (00 marks) Six questions to be answered. All questions carry equal marks. (6 0m). Four points P, Q, R, S lie on a straight level road. A train, travelling with uniform acceleration, passes point P with a constant speed of 0 m s and seconds later passes Q with a speed of 0 m s. The time taken by the train to travel from Q to R is twice the time it takes to travel from R to S. Its speed at S is 6 m s. Find (i) the uniform acceleration of the train (0) For PQ, u 0, v 0, t v u at 0 0 ( a)() 0 a a ms (ii) PQ, the distance from P to Q (0) For PQ, u 0, v 0, a v u as (0) (0) ()( s) s 4s 00 s 7 m 0. L.4/4_MS 4/ Page 4 of 48 exams

5 (iii) QS, the distance from Q to S (0) For QS, u 0, v 6, a v u as (6) (0) ()( s) 76 4s s 69 m (iv) RS, the distance from R to S. (0) For QS, v u at 6 0 ()( t) 6 t t s Time from Q to R is () s. For QR, u 0, a, t v u at v ( 0) ()() v 4 m s For RS, u 4, v 6, a v u as (6) (4) ()( s) s 4s 00 s m 0. L.4/4_MS / Page of 48 exams

6 . At a certain instant, a boat A is 96 m due north of another boat B. A A is travelling at a constant speed of 6 m s in the direction 0 west of south. 6 m s 0 B is travelling due west at a constant speed of 8 m s. 96 m Find (i) the velocity of A in terms of i and j (0) v A v A 6 sin0i 6cos0 j 8 m s B i 0 j (ii) the velocity of B in terms of i and j () v B 8i (iii) the velocity of A relative to B in terms of i and j (0) v AB v AB v AB va vb ( i 0 j) ( 8i ) i 0 j (iv) the magnitude and direction of the velocity of A relative to B (0) vab (0) 7ms v AB and 0 tan 04 tan θ v r AB 0 0. L.4/4_MS 6/ Page 6 of 48 exams

7 (v) the shortest distance between them in their subsequent motion. () Let M be the point of closest approach on the path of A relative to B. Shortest distance: p BM 96cos 96cos m A 96 θ θ v r AB p M B 0. L.4/4_MS 7/ Page 7 of 48 exams

8 . A particle is projected from a point O on horizontal ground with an initial speed of u m s at an angle of 60 to the horizontal. u A One second after being projected, the particle passes a point A, which is 0 m horizontally from the point of projection. O 60 0 m Find (i) the value of u When, t, s x 0 u cos 60 () 0 0 u u 40 (0) (ii) the height of A above the horizontal ground (0) The height of A above the horizontal at t s y 40sin 60 () (0)() 9 64 m (iii) the maximum height of the particle above the horizontal ground () The maximum height occurs when v 0 y 40sin 600t t t 464 The maximum height at t 464 s y 40sin 60 (464) (0)( 464) 60 m 0. L.4/4_MS 8/ Page 8 of 48 exams

9 (iv) the other time that the particle has the same vertical height above the plane as at A. () Let t be the times when the particle is at the height of 9 64 m. sy sin 60t (0) t tt t 464t9 64 t 464 t (464) 4()(9 64) t () t 0 (We want the greater time.) t 0 t 98s 0. L.4/4_MS 9/ Page 9 of 48 exams

10 4. A particle of mass 4 kg is connected to another particle of mass kg by a taut, light, inextensible string which passes over a smooth light pulley at the edge of a rough horizontal table. The coefficient of friction between the 4 kg particle and the table is. 4 kg The system is released from rest. kg (i) Show on separate diagrams the forces acting on each particle. (0) N a T μn 4 kg T kg a 4g g (ii) If =, find the common acceleration of the particles. (0) For the 4 kg mass: N 4g T N 4a For the kg mass: gt a Then: T (4 g ) 4 a 8g T 4a Adding and, 8g g 7a g 7a 0 ms ms a 0. L.4/4_MS 0/ Page 0 of 48 exams

11 (iii) Find the tension in the string. (0) From, 0 0 T 0 T T N 8.7 N 7 (iv) Calculate the time it takes the kg particle to fall metres. (0) u 0, a 0476, s s ut at (0)( t) (0 476) t 0 8t t 8 40 t 90s 0. L.4/4_MS / Page of 48 exams

12 . A smooth sphere A, of mass kg, collides directly with another smooth sphere B, of mass kg, on a smooth horizontal surface. A and B are moving in opposite directions with speeds of m s and 4 m s, respectively. A m s 4 m s kg B kg The coefficient of restitution for the collision is. Find (i) the speed of A and the speed of B after the collision (0) PCM: ()() ()( 4) () v () v... (0m) 7v v vv 7 NEL: v v ( 4)... (0m) 4 vv 8 vv Adding and, 7 7v ms v and v 7 6 v ms v... (0m) (ii) the percentage loss in kinetic energy due to the collision () KE before collision ()() ()( 4) 8 KE after collision () () 906 Loss in KE L.4/4_MS / Page of 48 exams

13 Percentage loss in KE % 8 0 0% (iii) the magnitude of the impulse imparted to B due to the collision. () Impulse 0 () ()( 4) Ns 667 Ns 0. L.4/4_MS / Page of 48 exams

14 6. (a) Particles of weight N, N, N and N are placed at the points (4, ), (p, ), (p, q) and (, q), respectively. The co-ordinates of the centre of gravity of the system are (6, ). Find (i) the value of p (0) (4) ( p) ( p) () 6 66 p p p 8 6p 48 p 8 (ii) the value of q. (0) ( ) ( ) ( q) ( q) q 4q 6 9q 7 9q q 0. L.4/4_MS 4/ Page 4 of 48 exams

15 (b) A quadrilateral lamina has vertices A, B, C and D. B The co-ordinates of the vertices are A(, ), B(, 9), C(8, ) and D(, ). Find the co-ordinates of the centre of gravity of the lamina. A C (0) B (,9) D 6 A (, ) C(8,) 9 6 D (, ) ABC ACD Area (9)(6) 7 (9)(6) 7 Centre of gravity 8 9, (,) 8, (4,)... (0m)... (0m) Let (x,y) be the centre of gravity of the lamina. Then (7)() (7)(4) x x and (7)() (7)() y 7 7 y Thus the co-ordinates of the centre of gravity are (, ). 0. L.4/4_MS / Page of 48 exams

16 7. A uniform rod, [AB], of weight 0 N is smoothly hinged at end A to a rough vertical wall. C The rod is held in a horizontal position by a light, inextensible string attached to point B on the rod and to the point C on the wall vertically above A. A θ B The string [ BC ] makes an angle with the rod [ AB ], as shown in the diagram. (i) Show on a diagram all the forces acting on the rod [ AB ]. (0) C F A N l 0 T θ l B... (0m) (ii) Write down the two equations that arise from resolving the forces horizontally and vertically. N T cos F T sin 0 (0) (iii) Write down the equation that arises from taking moments about the point B. (0) Let AB l. Then (0)( l) ( F)( l)...(m, m) 0. L.4/4_MS 6/ Page 6 of 48 exams

17 (iv) If the rod is in equilibrium and is on the point of slipping when tan, find the 4 coefficient of friction between the rod and the wall. (0) From the moments equation: 0 F F If tan, then 4 sin and 4 cos. Then and T 0 T T 4 N N 0 If the rod is on the point of slipping, then F N L.4/4_MS 7/ Page 7 of 48 exams

18 8. (a) A particle describes a horizontal circle of radius r metres with uniform angular velocity radians per second. Its speed is 8 m s and its acceleration is m s. Find (i) the value of () Speed 8 r 8...(m) and Acceleration r...(m) then ( r) 8 4rads...(m) (ii) the value of r. () Then r(4) 8 r m...(m) (b) A right circular cone is fixed to a horizontal surface. Its semi vertical angle is, where 8 tan and its axis is vertical. A smooth particle of mass 4 kg describes a horizontal circle of radius r cm on the smooth inside surface of the cone. r The plane of the circular motion is 0 cm 0 cm above the horizontal surface. α (i) Find the value of r. () r 8 tan α 0 r 6cm 06 m 0. L.4/4_MS 8/ Page 8 of 48 exams

19 (ii) Show on a diagram all the forces acting on the particle () R α 4g (m) (iii) Find the reaction between the particle and the surface of the cone. (0) Rsin α 4g 8 R 40 7 R 8 N (iv) Calculate the angular velocity of the particle. (0) R cos mr R (4)(06)ω ω ω 87 ω 6 ω rad s 4 0. L.4/4_MS 9/ Page 9 of 48 exams

20 9. (a) State the Principle of Archimedes. () Statement A solid piece of metal has a weight of 0 N. When it is completely immersed in water, the metal weighs N. Find (i) the volume of the metal (0) B = weight of water displaced =000(V)(0) V. 0 m (ii) the relative density of the metal, correct to two decimal places. () Weight of metal ρvg 0 ρ( 0 )(0) ρ ρ s 000 (b) An object consists of a hemisphere of diameter cm surmounted by a cone of diameter cm and height 8 cm. The relative density of the object is 0 6 and it is completely immersed in a tank or liquid of relative density. The object is held at rest with its axis vertical by a light, inextensible vertical string which is attached to the base of the tank. Find the tension in the string. [Density of water = 000 kg m ] (0) and B 00 (006) (006) (008) (0) B 89 W 600 (006) (006) (008) (0) W 4 T W B T 89 4 T 77 N 0. L.4/4_MS 0/ Page 0 of 48 exams

21 exams Pre-Leaving Certificate Examination, 0 Applied Mathematics Higher Level Marking Scheme (00 marks) Six questions to be answered. All questions carry equal marks. (6 0m). (a) A train travels a distance d from rest at one station to rest at another station. The train travels for the first part of its journey with a constant acceleration f. It then immediately decelerates to rest at the second station with a constant deceleration f. Show that the total time taken is d f f. () Let v be the maximum speed attained. Then v : f t v t f v : f t v t f T t t : v v T f f T v f f T v f f v d t t T Further answers overleaf 0. L.4/4_MS / Page of 48 exams

22 4: d Tv T d T. f f d T f f T d f f (b) A particle, P, starts from rest at a point A and moves with constant acceleration f in a straight line. A time T, after P starts from A, a second particle, Q, starts from A and moves in the same direction along the same straight line as P. Q moves with a constant speed of u. (i) Prove that Q will overtake P if u ft. (0) Let t Then and T. Let s and s be the distances travelled by P and Q in time t. s (0)( t) ( f)( t) s ft s ( u)( t T) The particles will be level when s s ft ut ut ft ut ut 0 u ( u) 4( f)( ut) t f t u u fut f u u fut t f 0. L.4/4_MS / Page of 48 exams

23 For Q to overtake P, this equation must have real solutions, i.e. u fut 0 u fut u ft b 4ac 0 (ii) Assuming Q does overtake P, i.e. that u ft, express in terms of u, f and T the length of time for which Q is ahead of P. () u u fut u u fut Let t and t. f f The length of time Q is ahead of P is u u fut u u fut t t f f u fut f 0. L.4/4_MS / Page of 48 exams

24 . (a) A man can swim at m s in still water. He swims across a river of width 60 metres. The river flows with a constant speed of m s parallel to the straight banks. He swims at an angle to the upstream direction but ends up going at an angle to the downstream direction. m s m s α θ 60 m (i) Show that tan θ = sin. (0) cos Let i and j be unit vectors in the direction of the river flow and perpendicular to the banks respectively. Then vr i v cosi sinj MR α β α θ θ and v M v MR v R cosi sin j i cos i sin j If this makes an angle with the downstream direction, then sin tan cos θ cos α sin α 0. L.4/4_MS 4/ Page 4 of 48 exams

25 (ii) Find the time taken for the man to cross by the shortest path. () Let be the angle between v MR sin sin sin sin and v M. Then by the Sine rule, For the shortest path, we require the largest possible value of. This is when 90 and sin then vm 4 and 4 sin Then the speed directly across the river is 4 sin 4 Thus the time to cross the river by the shortest path is 60 4 s 0. L.4/4_MS / Page of 48 exams

26 (b) Ship A is travelling with a constant speed of 0 m s in the direction 0 north of east. At midday, ship B is 0 km due east of ship A, and is travelling in a straight line with a constant speed of v m s. (i) Calculate the minimum possible value of v if B is to intercept A. (0) va (0cos0 ) i (0sin 0 ) j v i j A v r A 0 Let B travel in the direction north of west. Then v ( vcos ) i ( vsin ) j Thus B v BA v B v A ( vcosi vsin j) i j A ( vcos ) i ( vsin) j For interception, v BA must point due west, i.e. vsin 0 sin v 00 v r B v α v r BA 0 km B For this to be possible, sin v v Thus the minimum value of v for interception is. (ii) If v = 6, show that B can travel in either of two directions to intercept A, and find these directions, correct to the nearest degree. () If v 6, then interception occurs when 6sin 0 sin 6 sin 6 or L.4/4_MS 6/ Page 6 of 48 exams

27 . (a) A particle is projected from a point on a horizontal plane with speed m s at an angle to the horizontal. The particle then strikes a small target whose horizontal and vertical distances from the point of projection are 0 m and 0 m respectively. Find (i) the two possible values of tan () Given: for the same value of t, sx 0 and sy 0 sx 0 : cos. t 0 0 t 7cos 0 g 00 sy 0 : sin 0 7cos 49cos 0 tan 0sec 0 tan (tan ) tan tan 0 (tan )(tan ) 0 tan or tan (ii) the two possible times taken to strike the target. (0) If tan, then 4 and cos 0 0 0s 7 7. The time taken is. If tan, then cos. The time taken is 0 0 9s L.4/4_MS 7/ Page 7 of 48 exams

28 (b) A particle is projected up an inclined plane from a point O, with initial speed of m s. The line of projection makes an angle with the inclined plane and the plane is inclined at an angle of 4 to the horizontal. The plane of projection is vertical and contains the line of greatest slope. The particle is moving horizontally when it strikes the inclined plane at Q. (i) Show that tan. () v x Q v y 4 v r θ O 4 v cos gsin 4. t and v sin gcos 4. t x gt vx cos and vy sin Also, time of flight: s 0 y y gt sin. t 0 70 sin t g At the time of flight: vy tan 4 v v x y x gt v g 70 sin g 70 sin cos sin g g cos70sinsin 70sin cos 0sin sin 0 cos tan 0. L.4/4_MS 8/ Page 8 of 48 exams

29 (ii) Find OQ. (0) Then tan sin and 0 cos 0 Then time of flight 70 0 g 0 7 and OQ s x at time of flight OQ 0 g OQ 0 OQ 70 7 m 0. L.4/4_MS 9/ Page 9 of 48 exams

30 4. (a) Two particles of masses kg and 6 kg are connected by a light inextensible string passing over a fixed smooth pulley. Initially the two particles are at rest at the same horizontal level. The system is released from rest. The 6 kg particle takes seconds to strike horizontal ground. kg 6 kg Find (i) the initial height of the particles above the ground () By Newton s nd Law: a T g and 4a 4g T adding 6a g g a Let h be the height of the particles above the ground. Then s ut at g h (0)() () g h 0. L.4/4_MS 0/ Page 0 of 48 exams

31 (ii) the greatest height above the ground to which the kg mass rises. (0) After seconds, vu at g g v (0) () When the 4 kg mass strikes the ground, the kg mass is at a height of 4g h above the ground. Let h be the further height the kg mass rises to before coming to rest. Then with a g and v u as g u g (0) ( gh ) 4g gh 9 g h 9 The maximum height of the kg mass is 4g g 4g m (b) A smooth wedge, of mass m and slope 4, rests on a smooth horizontal plane. A particle of mass m is placed on the inclined face of the wedge. m The system is released from rest. Find the speed of the mass m relative to the wedge, when the speed of the wedge is 0 m s. 4 m () Wedge ( ma ) Nsin 4 N ma N ma N ma N N mg 4 a Further answers overleaf 0. L.4/4_MS / Page of 48 exams

32 Particle N Forces mg Accelerations a 4º 4º a mg a mg b a mg m N a mg m ma ma mg 4ma a g g a a mg m b g g b g b Wedge g u 0, v0, a. Then vu at g 0 t t g Particle: vu at g v (0) g v ms 0. L.4/4_MS / Page of 48 exams

33 . (a) A smooth sphere A, of mass m, collides directly with a smooth sphere B, of mass m which is at rest on a smooth horizontal table. The coefficient of restitution between the spheres ise. The line of centres of the spheres is at right angles to a smooth vertical cushion at the edge of the table. Sphere B then strikes the cushion and rebounds. The coefficient of restitution between sphere B and the cushion is e. Show that there will be no further impact between the spheres if m e ee em. () Collision : Two spheres mum (0) mv m v u m v 0 m v PCM: mv mv mu NEL: vv eu u m v Then also mv mv mu mv mv emu ( m m) v ( m em ) u m em v u m m mv mv mu mv mv emu ( mm) v ( m em ) u m( e) v u m m Collision : v v Spheres and cushion e v em ( e) u m m There will be no further collision if v v m em em( e) u u m m m m m em e m em m e m ee m em m ( e ee ) em 0. L.4/4_MS / Page of 48 exams

34 (b) Two smooth spheres, each of mass m and radius r, collide while travelling on a smooth horizontal plane. Before impact, the speeds of the spheres are u and 4u respectively, and the spheres are moving in the same direction along parallel lines, a distance r apart. 4u u r The coefficient of restitution between the spheres is. Find the angle between their directions of motion after impact, correct to the nearest degree. () From the diagram r sin r Thus 4 cos and tan. 4 α r r Let and be the directions of motion of the two spheres after impact. Before u α 6u 4u u α 4u u r j i r m m After u β x u θ y PCM () i : 6u 4u mx my m m x y 4u L.4/4_MS 4/ Page 4 of 48 exams

35 NEL () i 6u 4u : x y 6u x y... Adding and, 4u x 7u x and 7u y 4u u y then u tan 7u and u tan u 99 Then the angle between their directions of motion L.4/4_MS / Page of 48 exams

36 6. (a) A particle P is moving at a constant speed on the inner surface of a smooth sphere of radius r. The particle is describing horizontal circles the sphere. r below the centre of Prove that the speed of the particle is 6gr. () From the diagram, r sin r 0 Then r rcos 0 r r Then R : mg R mg Circular motion: mv R r r mv ( mg) gr v 6gr v 4 v 6 gr mg R θ r r O C r R R sin0 = R 0 R cos 0 = R 0. L.4/4_MS 6/ Page 6 of 48 exams

37 (b) A particle moves with simple harmonic motion in a straight line. It has velocities of 4 m s and m s when it is at distances of m and m respectively from the centre of the motion. (i) Find the amplitude and the periodic time of the motion. () v 4 when x : v a x 4 a... v when x : a 4... Dividing by : 4 a a 4 a a 4 a 4 a 4 4a 6a a a a m From, 4 and T s (ii) Calculate the least time taken for the particle to travel from a position of rest to a position where its velocity is m s. (0) v 0 when xa Let x acos t x cost v when x cost cos t t t 0s 0. L.4/4_MS 7/ Page 7 of 48 exams

38 7. (a) A uniform ladder, of weight W and length l, rests with its lower end, P, on rough horizontal ground. Its upper end, Q, is in contact with a rough vertical wall. Q At both P and Q, the coefficient of friction is. The ladder makes an angle of tan to the horizontal. Express in terms of l, the distance that a person of weight W can safely climb before the ladder begins to slip. () P tan tan tan Then : N F W : F N Limiting friction: F N F N Let x be the distance that the person can ascend. x N θ P F l W N l W F Q Taking moments about P: N.lsin F.lcos Wl.cos Wx. cos ln tanlf Wl Wx 4lN lf Wl Wx then F N F N 9 thus N N W 9 9N N 8W 0N 8W 9W N and W W F, N then W W 4l l Wl Wx ll l x 9l x 9l x 0. L.4/4_MS 8/ Page 8 of 48 exams

39 (b) Two uniform rods, AB and BC, each of length l are smoothly jointed at B. The weight of AB is W and the weight of BC is W. The rods stand in equilibrium with the ends A and C on rough horizontal ground, with each rod making an angle with the vertical. A B α α C The coefficient of friction between A and the ground is, while the coefficient of friction between C and the ground is. The angle is increased until both rods are on the point of slipping. Find (i) the value of. () Structure ABC : N N 8W : N N Moments about A: ( W)( k) ( W)( k) N ( N)(4 k) 8W 4N A 9 N W Then 7 N W and 7 9 W W 7 7 N W B W μn k k k k N C 0. L.4/4_MS 9/ Page 9 of 48 exams

40 (ii) the value of when the rods are about to slip. (0) Rod AB Moments about B ( W)( lsin ) N(lcos ) ( N)(lsin ) 7W 7W ( W )(tan ) () ( tan ) 6 7 l tan 7tan N α 7 4tan 7 A N tan tan 0. 7 l W B 0. L.4/4_MS 40/ Page 40 of 48 exams

41 8. (a) Prove that the moment of inertia of a uniform rod of mass m and length l about an axis through its centre perpendicular to the rod is Standard Proof Moment of mass element Moment of body Integral Deduce ml. (0) (b) A uniform rod AB of mass m and length l has a particle of mass m attached at a distance x 0 from A. The system is free to rotate about a horizontal axis through A perpendicular to the rod. A m x When the system makes small oscillations about the horizontal axis through A, the length of the equivalent simple pendulum is 4l. B (i) Express x in terms of l. () Let h be the distance from A to the centre of mass of the system. ( m)( x) ( m)( l) h m m x l h Let I be the moment of inertia of the system about the horizontal axis through A. m = m m = m A x = x x = l Then B I Irod Iparticle 4 I ml mx Given I 4l Mh 4 ml mx 4l x l ( m) Further answers overleaf 0. L.4/4_MS 4/ Page 4 of 48 exams

42 4 4 l l x ( x l ) 4l x 4lx 4l x 4lx x 4l 4l x (ii) If the system is released from rest with AB horizontal, find the speed of B when it is vertically below A. () When AB is horizontal, KE 0 4l l PE ( m) g 7l mg 7mgl When B is vertically below A, KE I 4 4l ml m 4 9 ml PE 0 PCE: KEPE KE PE 7mgl 4 ml 9 g l g l g l 0. L.4/4_MS 4/ Page 4 of 48 exams

43 Let v be the velocity of B at its lowest point. Then v r v ( l) g l v g 4l l v 6gl 0. L.4/4_MS 4/ Page 4 of 48 exams

44 9. (a) 7 cm of a liquid of relative density liquid of relative density. is mixed with V cm of another If there is no contraction of volume, and the relative density of the mixture is 9 find the value of V. () m V ρ s and V 0 9 Then sv s V s V V 6 6 ()(70 ) ()( V 0 ) (70 ) ( V 0 ) 77V 9 7V 89V 77 V 7 0 V V = 4 (b) A block of mass kg, and relative density 4, is held suspended by a string attached to a scale A. The block is completely immersed in 00 cm of a liquid of relative density contained in a cylindrical beaker of mass 0 7 kg. The beaker sits on another scale B. (i) If scale A registers x kg, find the value of x. () Let W be the weight and B be the buoyancy. and W g B sw L s W T B 0. L.4/4_MS 44/ Page 44 of 48 exams

45 ( g) 4 06g In equilibrium: T B W T 06 g g T g If scale A registers a mass of x kg, then x. (ii) If scale B registers y kg, find the value of y. (0) Weight of liquid: 6 000() 00 0 g 8g Let W be the total weight of the beaker and the liquid. Then W 07 g8 g g B W R In equilibrium: RW B g 06 g g If scale B registers a mass of y kg, then y. (iii) The radius of the beaker is 0 cm. Find the height, in cm, of the liquid in the beaker, correct to two decimal places. (0) Let V be the volume of the block. Then m 000sV 000(4 )V V m 00 cm Total volume of liquid and block cm Let h cm be the height of the liquid. Then (0) h 000 h 67cm. 0. L.4/4_MS 4/ Page 4 of 48 exams

46 0. (a) A particle moving in a straight line experiences an acceleration of 4 cos t cm s at time t seconds. At time t 0 the particle is at rest and has 6 a displacement of 44 cm relative to a fixed point O on the line. (i) Find the first positive time that the particle reaches the point O. (0) d x 4cos t d t 6 dv 4cos t dt 6 dv 4 cos tdt 6 v4 6sin t c 6 v 4sin t c 6 When t 0, v 0 : 04sin0 c c 0 The unique solution is v 4sin t 6 Then dx 4sin t dt 6 dx 4 sin tdt 6 x 4 6cos t d 6 x 44cos t d 6 When t 0, x cos 0 d d d 0 The unique solution is: x 44cos t 6 When x 0 : 0 44cos t 6 0. L.4/4_MS 46/ Page 46 of 48 exams

47 cos 0 6 t t 6 t s t 9.4 (ii) Show that the particle is moving with simple harmonic motion. () acc 4cos t 6 x acc 4 44 acc 6 x As this is in the form acc x the particle is moving with simple harmonic motion. (b) A particle moving in a straight line of mass m is acted upon by a force of m magnitude directed away from a fixed point O on the line, where x is the x distance of the particle from O. The particle starts from rest at a distance d from O. Show that the velocity of the particle tends to a limit of d. () d x m m d t x v dv d x x vdv x dx 4 c v x 4 c v. 4 x Further answers overleaf 0. L.4/4_MS 47/ Page 47 of 48 exams

48 v c 4 x v 0 when x d : 0 4 d c c 4 d Unique solution: v 4 4 x d v 4 4 d x In the limit as x, lim v lim x x 4 4 d x lim d x 4 x d. d 0. L.4/4_MS 48/ Page 48 of 48 exams

49 Notes: 0. L.4/4_MS 49/ Page 49 of 48 exams

50 Notes: 0. L.4/4_MS 0/ Page 0 of 48 exams

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