Which kinectic energy in an unreferenced system?

I wake up after an indeterminate period of hibernation. I'm in the dark deep space. No reference at all. I don't ear the hum of the engines. I can't know whether my starship moves inertially or it has been trapped by some force-field and been fully stopped.

The radar is warning about a collison. Some object is getting closer at speed $$v$$. Using the telescope I recognize the object: it's another starship whose mass is 10 times the mass of my starship ($$m_2=10m_1$$). My starship has a shield that can resist an impact of energy $$E=3\frac{1}{2}m_1v^2$$. Will I survive?

The thing is that if it's me who moves then my kinectic energy is $$E_1=\frac{1}{2}m_1v^2$$. But if it's the other body who moves then the energy is $$E_2=\frac{1}{2}m_2v^2\;=10 E_1$$ (because $$m_2=10 m_1$$). Which is the right energy? $$E_1$$? $$E_2$$? A combination of both?

• What does "My starship has a shield that can resist an impact of energy $E=3\frac{1}{2}m_1v^2$" mean? – nwolijin Mar 16 at 17:27
• @nwolijin It means that if I knew that my starship hits a wall (everything located in the Earth) my shield protects against damage because $E_1<E=3E_1$ – Ripi2 Mar 16 at 17:31
• In it's current form this question has a lack of focus. In the title there is the expression 'unreferenced system'. At the start there is a statement about 'no reference'. But the question in the second paragraph, about a impending collision, has no connection with the statements about 'unreferenced system'. That is, whether or not the system is unreferenced has no bearing on the answer to the question in the second paragraph. – Cleonis Mar 16 at 17:57
• @Cleonis I can solve basic collisions problems (energy and momentum are conserved) in a reference system where $v_1$ and $v_2$ are known. How can I express that the only velocity I know is the relative velocity between the two objects? – Ripi2 Mar 16 at 19:08

The kinetic energy is not Galilean invariant (it will have different values in different frames of reference). Fortunately, there is a reasonable way to answer your question.

Consider that, after colliding, the approaching ship stops to a halt (this is the case of an inelastic collision). It is easy to see that this is the condition that all kinetic energy in the center of mass frame of the system was dissipated (which would be the work of the shield).

The center of mass velocity (see https://en.wikipedia.org/wiki/Center-of-momentum_frame) is just $$v_{cm} = \dfrac{m_2 v}{m_1 + m_2},$$ such that the velocity of the approaching ship is $$\frac{m_1 }{m_1 + m_2}v$$, and the velocity of your ship is $$-\frac{m_2}{m_1 + m_2}v$$.

The kinetic energy of both ships in this reference frame will then be: $$\dfrac{1}{2}m_2 \left(\frac{m_1 }{m_1 + m_2}v\right)^2 + \dfrac{1}{2}m_1 \left(-\frac{m_2 }{m_1 + m_2}v\right)^2 = \dfrac{m_2 m_1 v^2}{2 (m_1 + m_2)} = \dfrac{10}{11} \dfrac{m_1 v^2}{2} < 3\dfrac{m_1 v^2}{2}$$.

You will survive :).

The reason why you obtain different results when looking at the different frames of reference, stems from the fact that (I think, I may be wrong) you forgot to take into account the change in velocity of the bodies during the collision. If you are at rest and the other ship collides into you, you gain velocity and, as a consequence, the relative velocity decreases, decreasing the amount of kinetic energy in your reference frame (on top of the decrease associated with the change of energy), try to do this calculation. There is a difference between the total kinetic energy, and the kinetic energy that can be exchanged (which is the center of mass kinetic energy).

The definition "It means that if I knew that my starship hits a wall (everything located on the Earth) my shield protects against damage because E_1<E=3E_1", assumes that the object with which you are colliding has infinite mass $$-$$ that is the definition of a wall in the context of physics.

• Using the center of mass seems the most reasonable way. But it's a big surprise to me that $E_t=\frac{10}{11}E_1<E_1<E_2$ – Ripi2 Mar 16 at 19:20
• I think it will always be smaller than the minimum of the two. That is the role of physics, to defy our intuitions xD – JGBM Mar 16 at 20:03

The amount of kinetic energy that is converted upon collision is determined by the relative velocity, and by the relative velocity only.

Here is how you can examine that:
Simplify to the case of perfectly inelastic collision. That is, the colliding objects fuse to a single object.

For the state before the collision you assign a kinetic energy to each of the objects, for the state after the collision there is the single fused object, hence a single kinetic energy.

Start with using the coordinate system that is co-moving with respect to the Common Center of Mass (CCM) of the two objects. In any collision the velocity of the CCM does not change. That is: if you use the coordinate system that is co-moving with the CCM then after the collision the fused object is motionless.

Then do that assessment again, this time with a coordinate system that has a velocity with respect to the CCM.

You will find in that case: while all three assigned kinetic energies will come out different compared to the first assessment, the amount of kinetic energy that is converted upon collision is always the same.

You can do any ratio of masses of object 1 and object 2, you can use any coordinate system that has a uniform velocity with respect to the CCM of object 1 and object 2; the difference in amount of kinetic energy before and after the collision is always the same.