Conservation of Energy in Collision Consider two cars of mass $m$ travelling towards each other at velocity $v$. A bystander (mass $m_b$) on the side of the road would calculate that cars have a total kinetic energy of $mv^2$, while a driver would see the other car travelling towards him at $2v$ and the bystander travelling towards him at $v$. The driver would calculate the kinetic energy of the system as $2mv^2 + 1/2 m_bv^2$. The two cars then collide, and all their energy is converted to sound, heat, light, kinetic energy in the flung off pieces of debris, etc, and the two cars are brought to a stop. The two drivers get out, exchange information, and for kicks, calculate the energy of the system again, and this time, all 3 parties agree that the energy of the system is 0.
The fact that two different reference frames would calculate two different energies for a system doesn't confuse me: what does confuse me is, where did the extra energy in the driver's reference frame go?
 A: Since the drivers' velocities change, their reference frames are not inertial. Therefore, energy does not need to be conserved in their frames.
To get an idea of what happens to the extra energy, consider a third driver, also with velocity $v$, who does not collide and continues with the same velocity after the collision. This third driver's reference frame is inertial and coincides with one of the other drivers before the collision. After the collision, they measure the other two drivers to each have velocity $-v$, thus leaving them with a total kinetic energy of $mv^2$. In both this reference frame and the bystander's frame, the kinetic energy change of the colliding cars is $-mv^2$.
A: You are being inconsistent in your calculations. In the second case you mention, in which one driver considers himself stationary, after the collision both of the vehicles and the pedestrian are travelling at v compared with the driver's initial speed, so they will consider their KE  to be not zero in that frame but 2 plus 1/22, so the loss of KE is 2 as in the first case.
