Find: (a) velocity of sphere A after the impact, vA2 [ft/s], (b) velocity of sph

Question

Find: (a) velocity of sphere A after the impact, vA2 [ft/s], (b) velocity of sphere B after the impact, vB2 [ft/s], and (c) loss of energy during the impact.

Two small spheres A and B are moving towards each other as indicated in the figure. The weights of spheres are WA=9 [lb] and WB=15 [lb]. The velocities of spheres immediately before the impact are vA1=6 [ft/s] and vB1=2 [ft/s]. Knowing that coefficient of restitution is e=0.8 find: (a) velocity of sphere A after the impact, VA2 [ft/s], (b) velocity of sphere B after the impact, vB2 [ft/s], and (c) loss of energy during the impact.

Explanation / Answer

Coefficient of restitution e = (vB2-VA2)/(vA1-vB1) = 0.8

given that vA1= 6 ft/s

vB1 = 2 ft/s

then vB2-vA2 = 0.8*(vA1-vB1) = 0.8*(6-2) = 3.2 ft/s

Formulas For vA2 = (WA-eWB)*vA1/(WA+WB)] - [WB(1+e)*vB1/(WA+WB)]

vA2 = [(9-(0.8*15))*6/(9+15)] - [15*(1+0.8)*2/(9+15)]

A) vA2 = -0.75-2.25 = -3 ft/s

vB2 = (WB-eWA)*vB1/(WA+WB)] - [WA(1+e)*vA1/(WA+WB)]

vB2 = [(15-(0.8*9))*2/(9+15)] - [9*(1+0.8)*6/(9+15)]

B) vB2 = 0.65-4.05 = -3.4 ft/s

C) loss in energy is (1/2)[(WA*WB)/(WA+WB)]*(vA1+vB1)^2*(1-e)^2

0.5*[(9*15)/(9+15)]*(6+2)^2*(1-0.8)^2 = 7.2 J

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