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

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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.

Jun 8th, 2015

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]

Jun 8th, 2015

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Jun 8th, 2015
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Jun 8th, 2015
Oct 18th, 2017
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