# 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?

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.

• This is a nice answer, to put it in other words. You should treat the collision in center of mass of the objects colliding, if the center of mass itself is moving, say you have a collision of two balls inside a train, then all the residual objects will still be moving with the train, so that kinetic energy is still present as kinetic energy. Commented May 28, 2017 at 13:27
• So basically, it is a mistake to analyze kinetic energy in collisions from the reference frame of any of the individual objects? Commented May 28, 2017 at 15:05
• @An_African_Ape No it isn't - as long as you do a proper "before" and "after" analysis the result should be the same. Commented May 28, 2017 at 16:49
• Sorry I actually phrased that question really poorly. I think what I meant was if we are considering the kinetic energy before and after the collision from the same frame. For the situation you described using the centre of mass frame, should the kinetic energy before the collision be calculated using the sum of the two masses? i.e, ((1/2)x6000kg)x(20m/s)^2=1200kj? Commented May 28, 2017 at 17:43
• @An_African_Ape It is usually easier to work in the CoM frame, but you can do the work in the lab frame (for instance) in order to show explicitly that you get the same results for the energy loss (something I have asked students to do in the past). Commented May 28, 2017 at 17:50