Further Mathematics 2, Nov/Dec. 2010

Further Mathematics Paper 2, Nov/Dec. 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 18
(a) Two forces (3i + 5j) and (-2i + 3j) act on a body of mass 2 kg and cause it to move. Find, correct to two decimal places, the magnitude of the: (i) resultant force acting on the body; (ii) acceleration of the body; (iii) Change in velocity, if the forces acted on the body for 5 seconds. (b) A uniform beam of length 100 cm and mass 1.5 kg is placed on a pivot which is 20 em from one end. A vertical force, T, is applied upwards 5 cm from the other end to keep the beam in equilibrium. Calculate the: (i) magnitude of T; (ii) reaction at the pivot. [Take g = 10 ms-2]

Observation
This question was reported to being unpopular among the candidates and poorly attended by majority of those who attempted it. In part (a), candidates were expected to obtain the resultant of the two forces as F = (3i + 5j) + (-2; + 3j) = i + 8j. /FI/=√1 + 8² = √6S = 8.06N. The acceleration of the body is obtained from the relation F = ma where F = 8.06N, m = 2kg. Therefore a = 8.06 = 4.03 ms–². Change in velocity = a x t = 4.03 x 5 = 20.15ms–². 2 In part (b), candidates were expected to represent the given information diagrammatically as shown. Taking moments about P, we have: 15 x 30 = T x 75 => T 15 x 30/75 = 6N. The total upward force = total downward force. Hence, R + T = 15 i.e R + 6 = 15 => R = 15 - 6 = 9N.