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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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Jun 8th, 2015
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