Q:

Missing energy in moving reference frame

Today, a fellow student and I had a discussion and we can't figure out where we're making a mistake.  
Suppose you have a car with $v_1$ = 10 m/s that's heading towards another car with $v_2$ = 0 m/s. The kinetic energy of the car is then $E_1 \propto v_1^2$ and $E_2 = 0$. Thus $E_{\text{sum}} = E_1 \propto v_1^2$. If I put myself in a moving reference frame, that starts in the middle of both cars with a velocity of $v_\text{ref} = \frac12 v_1$ = 5 m/s towards the still car, I perceive both cars moving with a velocity of $v'_1 = -v'_2$ = 5 m/s = $\frac12 v_1$. For the kinetic energy follows $E_{\text{sum}} = E_1 + E_2 \propto 2 \left(\frac12 v_1\right)^2 = \frac12 v_1^2$.
So it seems that in my reference frame, the collision is not as devastating, as it factually is. I know this can not be right, because energy is conserved and it should be conserved regardless of my reference frame, right? So, where is the hidden mistake?

A:

Bear with me, because after the numerical examples I'm going to try to get to the heart of your question.
If $car_1$ with mass M were traveling at $10m/s$ because its motor had burned some fuel to accelerate it to that velocity, it would have $50Mm^2/s^2$ (or 25|M| J) of KE.   If it collided with $car_2$ of mass M which was resting at $0m/s$, and you were a stationary observer, you would see the two cars move off together after the collision with a combined mass of 2M at $5m/s$, according to the rules of conservation of momentum, with 25|M| J of KE, and you would conclude that the other 25|M| J of KE had been transferred to thermal and sound energy in the collision.
If you were running toward $car_2$ at 5m/s and the same collision occurred, you would see each of the two cars as moving at $5m/s$, and calculate that they each had 12.5|M| J of KE, and you would perceive them as coming to a halt when they collided.   Again you would say that 25|M| J of KE had been transferred to thermal and sound energy.    The crash was the same.
If you were running toward $car_1$ at $10m/s$ you would see $car_1$ moving at $20m/s$ and calculate that it had 200|M| J of KE, and would see $car_2$ moving at $10m/s$ with 50|M| J of KE.   After the crash they would appear to be moving at $15m/s$, and with their combined mass of 2M would appear to have 225|M| J of KE.   So yet again you would say that 25|M| J of KE had been transferred to thermal and sound energy.    The crash was the same.
These are examples of what is meant by kinetic energy being reference frame dependent. The calculated pre-collision KE for each car can vary greatly with the chosen frame of reference, yet they predict the same result in objective reality.   All observers would see the same final effects on the cars.   So KE is dependent on the frame of reference in the context of mathematical models, for the purposes of those mathematical models.   And there are examples of such modeling which are more sophisticated than this one.   Yet they all work out like this one to be true to reality because they must.   And what is this reality?
It is that there is only a certain amount of energy in the universe, and if some of it is transferred to an object in the form of kinetic energy, then that's where it stays until it gets transferred out of the object to something else.   In all of the scenarios above, the kinetic energy was actually in $car_1$, and its amount was 50|M| J.   That is because 50|m| J of the energy which had been stored as chemical energy in a certain amount of fuel and oxygen was transferred to $car_1$.   That fact does not depend on the frame of reference.   Energy transfer is objective.    However, once the car is moving, observers in different reference frames will measure different velocities and calculate different values of KE.    But the frame of reference models worked, because total energy is conserved even if we don't know or care which car really had how much kinetic energy.  So you were correct when you said, ". . . energy is conserved and it should be conserved regardless of my reference frame, right?"   Yes, right.

