In a given reference frame where object 1 (with known mass and velocity) collides elastically with object 2 (with known mass and velocity), can we identify which object loses kinetic energy? Is it always the more massive one? The faster one?
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2$\begingroup$ possible duplicate of Newtons Cradle, Collision Theory $\endgroup$– John AlexiouCommented Jan 31, 2014 at 21:35
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$\begingroup$ To figure out which way the momentum transfer goes look at this answer physics.stackexchange.com/a/80906/392. Momentum transfers in a such a way as to keep the final relative velocity on the opposite direction from the initial relative velocity. $\endgroup$– John AlexiouCommented Jan 31, 2014 at 21:35
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$\begingroup$ @ja72 How does it relate to the energy transfer? $\endgroup$– MichaelCommented Feb 1, 2014 at 8:54
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2 Answers
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I'm reinterpreting your question Which one transfers energy? as Which one loses kinetic energy? I think this still captures what you're after, since the total kinetic energy $K_\text{tot}$ remains the same.
In a strict sense, there's no single answer to this question because it depends on which frame of reference from which you choose to measure kinetic energy.
For example, let's say you find out that object 1 loses kinetic energy while object 2 gains kinetic energy. This would be interpreted as object 1 transferring energy. That's fine, nothing wrong with that. But now imagine measuring things from the rest frame of object 1. In this frame, object 1 is at rest initially, and after the collision it's moving. That means it gained kinetic energy, implying object 2 transferred energy to object 1, in contrast to our initial interpretation. So by measuring or calculating the result in a new frame, a new answer to your question pops out.
However, a related question one could ask is: In a given reference frame where object 1 (with known mass and velocity) collides elastically with object 2 (with known mass and velocity), can we identify which object loses kinetic energy? Is it always the more massive one? The faster one?
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$\begingroup$ I understand that my question wasn't enough clear. It would be nice if you could write all the possible answers to this question. $\endgroup$– MichaelCommented Feb 1, 2014 at 8:55
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Since Energy and Speed depend on the reference frame you adopt, I will choose the Center of Mass (or Center of Momentum) frame to give a first idea of the problem. First of all let us consider two particles of proper mass $m_1$ and $m_2$ respectively. Thus our equations are: $$ E_1 + E_2 = E_1'+E_2',\\ p_1 = p_2 = p,\ \ p_1' = p_2' = p' $$ where the un-primed quantities refer to the system before the collision, and the primed ones to the situation afterwards.
$$ E_1 + E_2- E_1' =E_2'\\ (E_1 + E_2- E_1')^2 =E_2'^2\\ E_1^2 + E_2^2+ E_1'^2 + 2E_1E_2 -2E_1E_1' -2E_2E_1'=E_2'^2 $$ now, using the relativistic shell-mass relation $E^2 - p^2 = m^2$ (where $c=1$), we get $$ E_1^2 + E_2^2+ E_1'^2 + 2E_1E_2 -2E_1E_1' -2E_2E_1'=p'^2+m_2^2 = E_1'^2-m_1^2+m_2^2\\ E_1^2 +p^2 + 2E_1E_2 -2E_1E_1' -2E_2E_1' + m_1^2 = 0\\ 2E_1^2 + 2E_1E_2 -2E_1E_1' -2E_2E_1' = 0 $$ which is easily recognized as $E_1 = E_1'$.
This means there is no energy transfer at all, in the CM reference frame, but only momentum transfer, depending on the angle at which the particles are scattered: $$ |\mathbf{p}_1 - \mathbf{p}_1'| = \sqrt{p_1^2 + p_1'^2 - 2p_1p_1'\cos\theta}. $$
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$\begingroup$ That's very nice answer but if we take for example the following question :Mass of one object is equal to 0.5 kg and its speed is equal to 1 meter per second. Mass of object 2 is equal to 2 kg and its speed is equal to 0.5 meters per second. The objects moving toward one another. You can see that the kinetic energy of each object before the collision and after is different. So my question is: Why? $\endgroup$– MichaelCommented Feb 1, 2014 at 9:03
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$\begingroup$ That happens because you are not observing the phenomenon from the Center of Mass frame: your momenta aren't equal and therefore total momentum is not zero! $\endgroup$ Commented Feb 1, 2014 at 13:02