### Question Description

A railroad car of mass *M* moving at a speed *v*_{1} collides and couples with two coupled railroad cars, each of the same mass *M* and moving in the same direction at a speed *v*_{2}.

(a) What is the speed *v*_{f} of the three coupled cars after the collision in terms of *v*_{1} and *v*_{2}?

(b) How much kinetic energy is lost in the collision? Answer in terms of *M*, *v*_{1}, and *v*_{2}.

## Final Answer

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a)Mv1+2Mv2=3Mvf

**vf=(v1+2v2)/3**

b)kinetic energy before coupeled=1/2*M*v1^2

kinetic energy after coupeled=1/2*M*vf^2

losses=1/2*M(v1^2-vf^2)

=1/2*M(v1^2-((v1+2v2)/3)^2)

=1/2M(v1^2*9-v1^2-4v2^2-4v1v2)/3

=**1/6 M(8v1^2-4v2^2-4v1v2)**