OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary General Certificate of Education Advanced General Certificate of Education MEI STRUCTURED MATHEMATICS 76 Mechanics Monday MAY 006 Morning hour 0 minutes Additional materials: 8 page answer booklet Graph paper MEI Examination Formulae and Tables (MF) TIME hour 0 minutes INSTRUCTIONS TO CANDIDATES Write your name, centre number and candidate number in the spaces provided on the answer booklet. Answer all the questions. You are permitted to use a graphical calculator in this paper. Final answers should be given to a degree of accuracy appropriate to the context. The acceleration due to gravity is denoted by g ms. Unless otherwise instructed, when a numerical value is needed, use g = 9.8. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. You are advised that an answer may receive no marks unless you show sufficient detail of the working to indicate that a correct method is being used. The total number of marks for this paper is 7. This question paper consists of 5 printed pages and blank pages. HN/5 OCR 006 [K/00/6] Registered Charity 066969 [Turn over
(a) (i) Find the dimensions of power. [] In a particle accelerator operating at power P, a charged sphere of radius r and density its speed increased from u to u over a distance x. A student derives the formula x 8p r u 9P (ii) Show that this formula is not dimensionally consistent. [5] (iii) Given that there is only one error in this formula for x, obtain the correct formula. [] (b) A light elastic string, with natural length.6 m and stiffness 50 N m, is stretched between fixed points A and B which are. m apart on a smooth horizontal surface. (i) Find the energy stored in the string. [] A particle is attached to the mid-point of the string. The particle is given a horizontal velocity of 0 m s perpendicular to AB (see Fig..), and it comes instantaneously to rest after travelling a distance of 0.9m (see Fig..). r. r has A A.m.m 0ms - 0.9m.m B.m B Fig.. Fig.. (ii) Find the mass of the particle. [5] 76 June 006
(a) A particle P of mass 0.6 kg is connected to a fixed point by a light inextensible string of length.8 m. The particle P moves in a horizontal circle as a conical pendulum, with the string making a constant angle of 55 with the vertical. (i) Find the tension in the string. [] (ii) Find the speed of P. [] (b) A turntable has a rough horizontal surface, and it can rotate about a vertical axis through its centre O. While the turntable is stationary, a small object Q of mass 0.5 kg is placed on the turntable at a distance of. m from O. The turntable then begins to rotate, with a constant angular acceleration of. rad s. Let w rad s be the angular speed of the turntable. w rads - F O F.m Q Fig. (i) Given that Q does not slip, find the components F and F of the frictional force acting on Q perpendicular and parallel to QO (see Fig. ). Give your answers in terms of w where appropriate. [] The coefficient of friction between Q and the turntable is 0.65. (ii) Find the value of w when Q is about to slip. [5] (iii) Find the angle which the frictional force makes with QO when Q is about to slip. [] 76 June 006 [Turn over
A fixed point A is m vertically above a fixed point B. A light elastic string, with natural length m and modulus of elasticity N, has one end attached to A and the other end attached to a particle P of mass 5 kg. Another light elastic string, with natural length.5 m and modulus of elasticity N, has one end attached to B and the other end attached to P. (i) Verify that, in the equilibrium position, AP = 5m. [] The particle P now moves vertically, with both strings AP and BP remaining taut throughout the motion. The displacement of P above the equilibrium position is denoted by x m (see Fig. ). A m 5m P xm B Fig. (ii) Show that the tension in the string AP is ( x) N and find the tension in the string BP. [] (iii) Show that the motion of P is simple harmonic, and state the period. [] The minimum length of AP during the motion is.5m. (iv) Find the maximum length of AP. [] (v) Find the speed of P when AP =. m. [] (vi) Find the time taken for AP to increase from.5 m to.5 m. [] 76 June 006
The region bounded by the curve y = x, the x-axis and the lines x and x is rotated through radians about the x-axis to form a uniform solid of revolution. p 5 (i) Find the x-coordinate of the centre of mass of this solid. [6] From this solid, the cylinder with radius and length with its axis along the x-axis (from x to x ) is removed. (ii) Show that the centre of mass of the remaining object, Q, has x-coordinate. [5] This object Q has weight 96N and it is supported, with its axis of symmetry horizontal, by a string passing through the cylindrical hole and attached to fixed points A and B (see Fig. ). AB is horizontal and the sections of the string attached to A and B are vertical. There is sufficient friction to prevent slipping. A R S B 0 x Fig. (iii) Find the support forces, R and S, acting on the string at A and B (A) when the string is light, [] (B) when the string is heavy and uniform with a total weight of 6N. [] 76 June 006
76 Mark Scheme June 006 (a)(i) [ Force ] = M L T [ Power ] = [ Force ] [ Distance ] [Time ] = [ Force ] L T (ii) = M L T ( L ) ( L T ) ( M L [ RHS] = M L T = T [ LHS ] = L so equation is not consistent (iii) [ RHS ] needs to be multiplied by which are the dimensions of u 8π r u ρ Correct formula is x = 9P OR α β γ δ ) L T x = k r u ρ P β = 8π r u ρ x = 9P E cao 5 or [ Energy ] = M L T or [ Energy ] T For ( L T ) and ( M L ) Simplifying dimensions of RHS With all working correct (cao) SR... L = 8 π T, so inconsistent 9 can earn E0 RHS must appear correctly Equating powers of one dimension (b)(i) Elastic energy is 50 0. 8 = 8 J (ii) In extreme position, length of string is. + 0.9 ( = ) elastic energy is 50. ( = 7 ) By conservation of energy, 7 8 = m 0 Mass is.98 kg Treat use of modulus λ = 50 N as MR for. + 0.9 or.5 or allow for ( ) 50 0.7 Equation involving EE and KE 5
76 Mark Scheme June 006 (a)(i) (ii) Vertically, T cos 55 = 0.6 9. 8 Tension is 0.5 N Radius of circle is r =.8sin 55 ( =.9 ) Towards centre, v T sin 55 = 0.6.8sin 55 M Give for one error OR T sin 55 = 0.6 (.8sin 55 ) ω ω =.7 v = (.8sin 55 )ω or T = 0.6.8 ω Dependent on previous Speed is 5.67 ms (b)(i) Tangential acceleration is r α =.. F = 0.5.. = 0.78 N Radial acceleration is = 0.7ω N F = 0.5.ω r ω =.ω SR F = 0.78, F = 0. 7ω penalise once only (ii) Friction F = F + F Normal reaction R = 0.5 9. 8 About to slip when F = μ 0.5 9. 8 (iii) 0.78 F tanθ = F 0.78 = 0.7. Angle is.5 + 0.9ω = 0.65 0.5 9.8 ω =. cao 5 For LHS and RHS Both dependent on Allow for tan Accept 0.9 rad F θ = etc F
76 Mark Scheme June 006 (i) T AP = ( = 88 ) T BP =.5 ( = 75 ).5 T mg TBP = 88 5 9.8 75 so P is in equilibrium AP = OR (AP ) = (BP.5) + 5 9. 8 B.5 AP + BP = and solving, AP = 5 E (ii) Extension of AP is 5 x = x TAP = ( x) = ( x) Extension of BP is 7 + x.5 =. 5 + x T BP = (.5 + x) = 9(.5 + x).5 (iii) d x ( x) 5 9.8 9(.5 + x) = 5 dt d x = 9x dt π Motion is SHM with period = π = 0.898 s ω 7 (iv) Centre of motion is AP = 5 If minimum value of AP is.5, amplitude is.5 Maximum value of AP is 6.5 m (v) When AP =., x = 0. 9 Using v = ω ( A x ) v = 9(.5 0.9 ) Speed is 8. m s OR x =.5sin 7t When x = 0.9, 7t = 0.65 ( t = 0.099) v = 7.5cos 7t = 0.5 cos(0.65) = 8. 0 E E Give for one tension correct Equation of motion involving forces d x Obtaining = ω x ( + c) dt Accept π 7 Accept ± 8. or 8. or x =.5 cos 7t or 7 t = 0.97 ( t = 0.5) or v = 7.5sin 7t = ( ) 0.5sin(0.97)
76 Mark Scheme June 006 (vi) x =.5 cos 7t When.5cos 7t = 0. 5 Time taken is 0.76 s For cos( 9 t ) or sin( 9 t) or x =.5sin 7t above can be awarded in (v) if not earned in (vi) or other fully correct method to find the required time e.g. 0.00 0. or 0. 0.09 Accept 0.7 or 0.8
76 Mark Scheme June 006 (i) π y π x y dx = = dx π x dx [ π x ] = 7.5π [ π ] ( π ) = π x dx = x = π x = 7.5π =.8 (ii) Cylinder has mass π ρ Cylinder has CM at x =. 5 (.5π ρ) x + (π ρ)(.5) = (7.5π ρ)(.8) x = (iii)(a) Moments about A, S 96 = 0 S = 6 N Vertically, R + S = 96 R = N (B) Moments about A, S 96 6.5 = 0 Vertically, R + S = 96 + 6 R = 5 N, S = 67 N OR Add N to each of R and S R = 5 N, S = 67 N A E 6 5 π may be omitted throughout Or volume π Relating three CMs ( ρ and / or π may be omitted) or equivalent, e.g. (7.5π ρ)(.8) (π ρ)(.5) x = 7.5π ρ π ρ Correctly obtained Moments equation or another moments equation Dependent on previous Moments equation Both correct Provided R S Both correct