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

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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}.
v_{f}  = 
(b) How much kinetic energy is lost in the collision? Answer in terms of M, v_{1}, and v_{2}.
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Okay so we know that momentum is going to be conserved and because the cars stick together the collision must be an inelastic collision. So let's now use the conservation of momentum.
momentum before collision = momentum after collision
M*v_1+ 2*M*v_2 = (M+2M)*v_final
Solving for v_final yields:
M*(v_1+2*v_2) = 3M*v_final
v_final = (v_1+2*v_2)/3
b) Now we need to look at how much energy is lost. So let's find the energy (in this case all of there energy is kinetic energy) before and after the collision. Before:
1/2*M*(v_1)^2 + 1/2*2*M*(v_2)^2 = 1/2*M*(v_1)^2 + M*(v_2)^2
Now let's look at the amount of energy after the collision:
Energy = 1/2*3*M*(v_final)^2
Substituting for v_final yields:
3/2*M*((v_1+2v_2)/3)^2
Now we expand either with bionomial expansion or with FOIL:
3/2*M*((v_1)^2 +2*v_1*2*v_2 + (2*v_2)^2)/9 =
1/6*M*(v_1)^2 + 1/6*M*4*(v_1)*(v_2) + 1/6*M*4*(v_2)^2 =
1/6*M*(v_1)^2 + 2/3*M*(v_1)*(v_2) + 2/3*M*(v_2)^2
Now we have to calculate the difference so we subtract final energy from the initial energy:
1/2*M*(v_1)^2 + M*(v_2)^2  (1/6*M*(v_1)^2 + 2/3*M*(v_1)*(v_2) + 2/3*M*(v_2)^2)
1/3*M*(v_1)^2 + 1/3*M*(v_2)^2  2/3*M*(v_1)*(v_2)
But now we can factor this (notice that it is a perfect square):
1/3*M*(v_1+v_2)^2
Notice that because v_1 and v_2 are both in the same direction that (v_1+v_2)^2 > 0 and there is a net loss in energy.
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