Further Mathematics 2, May/June. 2010

Further Mathematics Paper 2, May/June. 2010

Questions:	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	Main

General Comments
Weakness/Remedies
Strength

Question 16
(a) Two ships M and N, moving with constant velocities, have position vectors (3 i + 7 j) and (4 i + 5 J] respectively. If the velocities ofM and N are (5 i + 6)] and ( i+ 3 J] and the distance covered by the ships after t seconds are in meters, find i) MlJ (ii) lNfNI, when t = 3 seconds. (b) A particle is acted upon by forces F 1 = 5 i + pj, F2 = q r + j, F3 = -2p i + 3 j and F4 = -4 i + qj, where p and q are constants. If the particle remains in equilibrium under the action of these forces, find the values ofp and q. (b) A particle is acted upon by forces F 1 = 5 i + pj, F2 = q r + j, F3 = -2p i + 3 j and F4 = -4 i + qj, where p and q are constants. If the particle remains in equilibrium under the action of these forces, find the values ofp and q.
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Observation
Candidates' performance was reportedly very poor in this question. Teachers are encourage to emphasize this area of the syllabus. In part (a), candidates were expected to recall that displacement = velocity x time. Thus for N, new position = old position + vt = (4; + 5f) + t (2i + 3j) = (4+2t)i + (5+3t)j, similarly M = (3i + 7j) + (5i + 6J)t = (3 + 5t); + (7+6t)j. Hence lJN = [(4 + 2 t)i + (5+3t)j]- [(3+5t)i+(7+6t)J1 = (1-3t)i + (-2 - 3t)j. When t = 3,IlJN;1 = .J(-8)2 + (-11)2 = "'185 = 13.60 m. In part (b), since the system is in equilibrium, then Fl+F2+F3+F4 == 0 i.e. (5+q - 2p - 4); + (p + 1+3+ q)j = 0. Thus, q - 2p + 1 = 0 and p + q + 4 = 0. Solving these two equations simultaneously gave p = -1 and q = -3.