Question

4.

Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the same weight, has been pushed by dock workers and hits car B with a velocity of 1.75 m/s. Knowing that the coefficient of restitution is 0.8 between B and C and 0.5 between A and B. determine the velocity of each car after all collisions have taken place (Round your answer to three decimal places.) m/s - The velocity of car A is The velocity of car B is The velocity of car O is mis ms

Answer 1

briven two identical Cars ; A&B > MA=MB Car ima-meam of a different style but Same wength is > velocity of Cape C => Ve 31-75 m/s » Coeff of restitution between B&C 3 esc between AGB = CAB=0.5 ů first Collision between B&C Vg! Vd Vigro 42=1.750/ B C 3 - Apply conservation of momentum mub tmvc=mvightinue Voltvcl = 1.55 s) using relative velocity Vc - Vo = eac (VB-Vc) → V-VE' - 0-3 (0-1-75) =) Ve-ve' 1.4: 2 Solve {2. Ve'= 0.175m/s() { Ve' 1.575

Gis comparing Vo & ve are conclude that care Collides with Car A TV Vazo Vr3= 1 SISMI A 3 A 1) *mvat mug = mua thro" 1-575:- VA' + V6" >> Valtve" 1.575 -> using relative velocity Ve" - VAR s las (va-ve') - 0.5(0-1.575) =) Vo" -VA' Vor-vá = -7825. 2 Solve ② & 6 11 Vo = 0.39375F { Va' = 1-8125 m/s f VA'>Vg"> vel Conclusion do Velocity of Car A = 1.18125m/s a 6 Velocity of Car 8 = 10.39375m/st lis, velouts of cor = (0.175m/s Answers Car c =

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