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Question 3 (a) A 60 kg block is at rest on a smooth horizontal surface as shown in Fig. 23a. A force of 90 N acting at an angle of 35° is applied to the block for 7s. Determine the final velocity and the normal force which the smooth surface exerts on the block during the time interval F-90N u=0 Fig. 23a (5 marks) (b) A ball of mass 625 g hits the ground with the following velocity (Fig. 23b). The coefficient of restitution is 0.88. 15 m/s Lecture & Notes - 2016 Coments and principle of Fig. 23a (5 marks) (b) A ball of mass 625 g hits the ground with the following velocity (Fig. Q3b). The coefficient of restitution is 0.88. 15 m/s 20 Fig. 236 Given that the time of collision is 0.3 s, find the average contact reaction force acting on the ball during the impact (10 marks)

Answer 1

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(a)NFSINS > F-90N 56 FC6531 Efy = -0 (60x9-8) N N+F sims - Cox 9.81 - 0 N = GoX 9-81 - 9ox Sin38 = $36.97 N Hence, normal force exerted by smuth 149444 i 536-97 N. { Fu = man F (of 3r = 60(2) 90 (035 = 6019) .:19 = 1.2287 mis? V = U + 4 + V = 0 +(1-2257 x 7) 8.6 m/s Ans. Hence; Velocity after 7s is 8.6 m/s. (3) (b) 21 Ersand is war vy = is simai = 5:13 mls (1) e = 0.88 Vol = 0.88x5-13 = 4.5146 mis () (0.625) { 4.514-6-8-1039 impulse memerdum(P) = m(vyl – vy) . P = 6.0279 kg mis we know that = my-V) {f.dt favxt = Р FAT 60279 = 20.09 N t (0-3) ta)