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GCE JUN 2009 : AS, M4: Mechanics 4

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ADVANCED General Certificate of Education 2009 Mathematics assessing Module M4: Mechanics 4 AMM41 Assessment Unit M4 [AMM41] WEDNESDAY 17 JUNE, MORNING TIME 1 hour 30 minutes. INSTRUCTIONS TO CANDIDATES Write your Centre Number and Candidate Number on the Answer Booklet provided. Answer all six questions. Show clearly the full development of your answers. Answers should be given to three significant figures unless otherwise stated. You are permitted to use a graphic or a scientific calculator in this paper. INFORMATION FOR CANDIDATES The total mark for this paper is 75 Figures in brackets printed down the right-hand side of pages indicate the marks awarded to each question or part question. Answers should include diagrams where appropriate and marks may be awarded for them. Take g = 9.8ms 2, unless specified otherwise. A copy of the Mathematical Formulae and Tables booklet is provided. Throughout the paper the logarithmic notation used is ln z where it is noted that ln z loge z 4166 Answer all six questions. Show clearly the full development of your answers. Answers should be given to three significant figures unless otherwise stated. 1 A guitar has six strings each of which produces a fundamental frequency. The frequency f depends on the tension P in the string, the length l of the string and the mass per unit length of the string. The frequency is related to the other variables as follows: f = k P l where k is a dimensionless constant. (i) Use the method of dimensions to find and and show that = 1 [7] The fundamental frequency f1 of the first string S1 is four times that of the sixth string S6 The mass per unit length of S1 is 1 and that of S6 is 6 Both strings have the same tension and length. (ii) Show that 6 = 16 1 4166 [3] 2 [Turn over 2 Triangle ABC is isosceles with equal sides AB and AC. Angle BAC is 2 , where cos = 0.8 Forces of magnitude 10 N, 10 N and 30 N act along the sides AB, AC and BC respectively as shown in Fig. 1 below. The mid point of BC is M. A 2 10N 10 N M B C 30 N Fig. 1 (i) Find the resultant of the two forces of magnitude 10 N, stating its magnitude and two points on its line of action. [3] (ii) Hence find the magnitude of the resultant of the three forces and identify one point on its line of action. [3] The distance AB is 0.85 m. The distance from B to the line of action of the resultant is d. (iii) Find d. 4166 [5] 3 [Turn over 3 Take g = 10 m s 2 in this question. The Ratzan a large rodent recently discovered on a tropical island swings from tree to tree at the end of a long vine. The weight of the Ratzan is 200 N. A Ratzan was photographed at A moving at 16 m s 1 in a vertical circle holding the end of a 10 m vine that made an angle of 60 with the downward vertical. The Ratzan, R, can be modelled as a particle on the end of a light inextensible rope whose other end is fixed at C, as shown in Fig. 2 below. C 60 A 16 m s 1 Fig. 2 (i) Find the tension in the vine when the Ratzan was at A. [5] After leaving A the Ratzan swung the taut vine through 180 to B as shown in Fig. 3 below. B 60 C 60 A 16ms 1 Fig. 3 (ii) Find the tension in the vine when the Ratzan was at B. 4166 4 [8] [Turn over 4 Two particles of masses 3m and 2m are moving towards one another along a straight horizontal track. Before they collide, their speeds are 3u and 2u respectively. Afterwards their velocities are v1 and v2 as shown in Fig. 4 below. Before 2u 3m After 3u 2m v1 v2 Fig. 4 The coefficient of restitution between the particles is e. (i) Find v1 and v2 in terms of e and u. [7] (ii) Show that the loss in kinetic energy caused by the collision is 15mu2(1 e2) (iii) If 37.5% of the initial kinetic energy is lost in the collision, find e. 4166 5 [4] [3] [Turn over 5 2 a3 A solid hemisphere has radius a, density and volume V = 3 (i) Write down an expression for its mass in terms of a and . [1] A solid hemisphere is produced by rotating the area bounded by the curve y = (a2 x2), between x = 0 and x = a, and the x-axis through 2 radians about the x-axis as shown in Fig. 5 below. y a 0 x Fig. 5 (ii) Find, in terms of a, the x coordinate of the centre of mass of the hemisphere. [7] An ornamental bowl is made by removing a solid hemisphere of radius 15 cm from a larger one of radius 20 cm. (iii) Find the distance of the centre of mass of the bowl below the rim of the bowl. 4166 6 [6] [Turn over 6 At a certain instant the artificial satellite Xeo, X, lies directly between the Earth, centre E, and the Moon, centre M. The distance EM is d metres. The resultant of the gravitational pulls of the Earth and Moon on Xeo is zero at the point X where EX = fd and 0 f 1, as shown in Fig. 6 below. M d X fd E Fig. 6 The masses of the Earth and Moon are ME and MM respectively. The universal gravitational constant is G. (i) Show that ME f2 = MM (1 f )2 [5] (ii) Given that ME = 81MM, find the value of f. [3] Some time later the Sun, centre S, Moon, Xeo and Earth are in line and in that order. The distance EX = 0.75ME. The masses of the Sun, Moon and Earth are 1.99 1030 kg, 7.35 1022 kg and 5.98 1024 kg respectively. The distance SX is 1.49 1011 m and ME is 3.84 108 m. The mass of Xeo is 500 kg. The value of G is 6.67 10 11 m3 kg 1 s 2 (iii) Find the resultant of the gravitational forces of the Sun, Moon and Earth acting on Xeo. [4] With the configuration of Sun, Moon and Earth as above and the Sun s gravitational field included, the position of zero resultant gravity between the Earth and Moon would be closer to the Earth. (iv) How do the results of (ii) and (iii) show this? 4166 7 [1] [Turn over S 1/08 938-006-1 [Turn over

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Additional Info : Gce Mathematics June 2009 Assessment Unit M2 Module M2 : Mechanics 4
Tags : General Certificate of Education, A Level and AS Level, uk, council for the curriculum examinations and assessment, gce exam papers, gce a level and as level exam papers , gce past questions and answer, gce past question papers, ccea gce past papers, gce ccea past papers

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