If kinetic energy depends on your reference frame then which frames kinetic energy is dissapated during a collision to cause damage etc? Say you have an object such as a car with mass 4,000kg moving at a speed of 20m/s relative to the ground in a straight line towards another object with, say, 2,000kg which is stationary on the ground which of course gives it 0j of kinetic energy. From the reference frame of the ground or the 2,000kg object the car is moving towards it with a kinetic energy of:
((1/2)x4000kg)x(20m/s)^2=800kj.
From the reference frame of the car it is not moving and so has 0j of kinetic energy with the 2,000kg object moving towards it at the same speed of 20m/s resulting in it having:
((1/2)x2000kg)x(20m/s)^2=400kj
So when they collide, how much kinetic energy will be dissipated as damage and sound etc: the 800kj, 400kj, or perhaps both?
 A: After the collision, assuming for a moment it is perfectly inelastic, the fragments in the center of mass frame will have no residual energy. However, in any other frame the combined object will have a final velocity and thus residual energy.
The conclusion is that the kinetic energy in the center of mass frame is what gets dissipated. Any other energy from "before" will still be present as kinetic energy "after".
Example calculation using the numbers in your question:
In the "ground" frame of reference - energy before = $$\frac12 m_1 v_1^2 = \frac12 \cdot 4000 \cdot 20^2 = 800~kJ$$
After (inelastic) collision, velocity (from conservation of momentum) is $$v_2 = v_1\cdot \frac{m_1}{m_1+m_2}$$
and kinetic energy after is $$\frac12 (m_1+m_2) v_2^2 = \frac12 \frac{m_1^2}{m_1+m_2} v_1^2$$
The energy lost is $$\Delta E = \frac12 \left(m_1 - \frac{m_1^2}{m_1+m_2}\right)v_1^2\\
= \frac12 \frac{m_1 m_2}{m_1+m_2} v_1^2$$
The quantity $\frac{m_1m_2}{m_1+m_2}$ is called the reduced mass and shows up frequently in two-body problems. It turns out that this is the energy lost in the collision regardless of the frame of reference.
Let's assume we have a frame of reference with velocity $v_f$. Then the energy before is
$$E = \frac12 m_1 (v_1 - v_f)^2 + \frac12 m_2 v_f^2$$
After the collision the velocity of the two particles is now $v_2+v_f$ (for the $v_2$ I had before). If you work through the math you will get the same energy lost.
