 # A railroad car of mass M moving at a speed v1 collides and couples with two coup

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### Question Description

A railroad car of mass M moving at a speed v1 collides and couples with two coupled railroad cars, each of the same mass M and moving in the same direction at a speed v2.

(a) What is the speed vf of the three coupled cars after the collision in terms of v1 and v2?
 vf =

(b) How much kinetic energy is lost in the collision? Answer in terms of Mv1, and v2. Student has agreed that all tutoring, explanations, and answers provided by the tutor will be used to help in the learning process and in accordance with Studypool's honor code & terms of service. momentum will be conserved after collision. Therefore, p(i) = M*v1 + M*v2 +M*v2 and p(f)=(M+M+M)*Vf

Vf =(M*v1+M*v2+M*v2)/(3M) = (v1 +2*v2)/3, since M*v1+M*v2+M*v2 is the momentum after collision.

b) the kinetic energy:

K(i)=1/2*M*v1^2 +1/2*M*v2^2 +1/2*M*v2^2

K(f)= 1/2*M/3*[v1^2 +4*v1*v2+4*v2^2]

therefore energy lost= K(f)-K(i)

=1/2*M*v1^2 +1/2*M*v2^2 +1/2*M*v2^2-{1/2*M/3*[v1^2 +4*v1*v2+4*v2^2]}

=M/3*[-v1^2 + 2*v1*v2 -v2^2] ZnguPnyphyngbe (47)
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