Consider a model for two stationary rail cars, B and C, of masses mB and mC, con

Question

Consider a model for two stationary rail cars, B and C, of masses mB and mC, connected together by a spring coupling of spring constant k. A third car A, of mass mA, is traveling along the frictionless track at a constant speed vA until it collides with the two coupled cars. Model the three rail cars as point masses, arranged as in Figure 6.39. Assume that the coupling is broken on car A and that the collision of car A with car B is perfectly elastic (e = 1). What is the speed of the center of mass of cars B and C after the collision?

Explanation / Answer

velocity of B just after the collision will be balancing momentum mb*Vb+ma*va1=ma*va [va1 is velocity of a after collision]............1) conserving energy 1/2*mb*vb^2+1/2ma*va1^2=1/2ma*va^2....................................2) From these 2 equations vb and va1 can be found out. initial velocity of the center of mass of B and C is mb*vb/(mb+mc) As the net force after the collision on system containing b and c is 0 velocity remains unchanged and final velocity of center of mass is mb*vb/(mb+mc) vb can be found out from 1 and 2

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