# Really basic mechanics and Galilean Relativity question

Consider two solid objects: A and B.

System 1:

A        <----- B
10m/s


System 2:

A ----->        B
10m/s


They are identical except for the reference frame.

Common relativity tells us they are equivalent and will evolve in the same way, right?

Now add masses: A weights 1kg and B weighs 10kg.

In system 1, B will hit A with 500J of kinetic energy. In system 2, A will hit B with 50J of kinetic energy.

So... the energy involved in the impact (and effects such as how far debris is flung) depends on what reference frame I'm in? This doesn't seem right.

How do we resolve this?

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As it stands your system isn't completely constrained in the final state. Momentum will be conserved for sure (net external force is zero) but until you say whether kinetic energy is conserved or what one of the final velocities will be you can't solve it. So I suggest picking some final velocities and prove to yourself that things work out. –  DJBunk Jul 9 '12 at 16:20

I think that your intuition is leading you astray here; KE and momentum are frame dependent quantities. To get a handle on, let's say, the violence of a collision, use the center of momentum (COM) frame of reference.

Let's use an absurd example to see how your intuition might mislead you. Consider the collision of some low-mass particle with an arbitrarily massive one. In the massive particle's frame, the low-mass particle approaches at a very slow speed and thus, has little KE and momentum. The collision is barely a kiss.

However, in the low-mass particle's frame, the KE and momentum of the massive particle is arbitrarily large!

The resolution to this "paradox" is, as mentioned before, to go to the COM frame. In this frame, the particles have equal and opposite momentum. In the example above, the COM frame effectively is the massive particle's frame.

In the COM frame of your example, particles A & B have KE of 41.3J and 4.13J respectively. Note that the total energy is less than the 50J in reference frame A. In fact, it is the case that the total energy of the system is minimum in the COM frame.

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