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

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Science

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?

(b) How much kinetic energy is lost in the collision? Answer in terms of Mv1, and v2.

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Explanation & 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)

Please let me know if you need any clarification. I'm always happy to answer your questions.


Anonymous
Really great stuff, couldn't ask for more.

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