2 fts 2 4 lb 322 fts 2 0062112 lb s 2ft 0124224 lb s

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Unformatted text preview: energy: 1 2 mA v A 2 T0 V0 (2)(0.40192) (3)(1 cos 30 ) 0.40192 ft 0.80385 ft lb 0.12422 ft lb 0 1 2 (0.062112) v A 2 2 0.031056 v A T1 V1 2 0.12422 0.80385 0.031056 v A 2 vA 2 29.884 ft /s vA 0 2 5.4666 ft/s Analysis of the impact. Use conservation of momentum together with the coefficient of restitution e 0.8. Note that the ball rebounds horizontally and that an impulse Tdt is applied by the rope. Also, an impulse Ndt is applied to B through its supports. PROBLEM 13.188 (Continued) Both A and B. Momentum in x-direction: m A (v A ) x 0 (0.062112)(5.4666) Coeffi...
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This note was uploaded on 12/01/2012 for the course ENGR 240 at Canada College.

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