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

Physics
Tutor: None Selected Time limit: 1 Day

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

Thank you for the opportunity to help you with your question!

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. 

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 8th, 2015

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