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Imagine a ball ($m= 1\,{\rm kg}$) moving at a velocity $2\,{\rm m}/{\rm s}$ towards a wall. When it hits the wall, it suddenly stops, thereby liberating all its ${\rm KE}$ as heat. Here, the initial kinetic energy, ${\rm KE}_I = (1/2)mv^2 = 2\,{\rm J}$, and final ${\rm KE}$ is obviously zero. So heat liberated (${\rm KE}_F - {\rm KE}_I$) equals $2\,{\rm J}$.

Now, suppose I hop on a car moving at $2\, {\rm m}/{\rm s}$ towards the oncoming ball from the direction of the wall. So for me, the ball is initially moving at $4\, {\rm m}/{\rm s}$, and after impact, it is moving at $2\, {\rm m}/{\rm s}$ (relative velocity). In this case, the ${\rm KE}_I=8\,{\rm J}$, and the ${\rm KE}_F=2\,{\rm J}$. In this case the heat liberated (${\rm KE}_F - {\rm KE}_I$, due to conservation of energy) equals $6\,{\rm J}$.

So, is this observation correct? If it is, then how does it make any sense? And if not, then what is wrong with the above approach?

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Initially your ball has some energy ($2J$) and some momentum ($2Ns$). And the wall has some energy ($0J$) and some momentum ($0Ns$). And there is some internal energy, $U=U_0,$ the thing that heat increases.

Afterwards the ball has some energy ($0J$) and some momentum ($0Ns$). And the wall has some energy ($0J$) and some momentum ($2Ns$). And there is some internal energy, $U=U_0+2J,$ the thing that heat increased.

Note that the wall has some momentum but no energy. Why? Because it has an infinite mass and $K.E.=p^2/(2m).$ And it has infinite mass because that's the only way it can acquire the momentum from the ball without itself changing its velocity.

So in the moving frame everything is different.

Initially your ball has some energy ($8J$) and some momentum ($4Ns$). And the wall has some energy ($\infty J$) and some momentum ($\infty Ns$).

Afterwards the ball has some energy ($2J$) and some momentum ($2Ns$). And the wall has some energy ($\infty J$) and some momentum ($\infty Ns$).

Energy and momentum are conserved in both frames, but it isn't helpful when you have infinitely massive objects.

What if you give the wall some finite mass? If you have a ball of mass $m$ and a wall of mass $M$ and the initial momentum is $p$ then there is an initial energy $p^2/(2m).$

Then after the collision the final velocity is $v_1=p/(m+M).$ This means the kinetic energy of the ball is $\frac{mp^2}{2(m+M)^2}$ and the kinetic energy of the wall is $\frac{Mp^2}{2(m+M)^2}.$ So the total kinetic energy afterwards is $\frac{p^2}{2(m+M)}.$

So the change in kinetic energy is $\frac{p^2}{2m}-\frac{p^2}{2(m+M)}$ which equals $\frac{Mp^2}{2m(m+M)}.$ This is how much energy goes into heat. And before we go on this result has a great physical interpretation in terms of what happens when M goes to infinity. You had an initial energy of $p^2/(2m)$ and $M/(m+M)$ of it goes into heat, the rest goes into kinetic energy of the wall.

Now we will look at the frame that moves an additional speed $u$ relative to the wall.

The initial momentum of the ball is $p+um$ and its initial kinetic energy is $(p+um)^2/(2m).$ And the wall has an initial momentum of $uM$ and its initial kinetic energy is $(uM)^2/(2M).$ So the total initial kinetic energy is $\frac{(p+um)^2}{2m}+\frac{(uM)^2}{2M}$ which equals $\frac{M(p+um)^2+m(uM)^2}{2mM}.$

Then after the collision the final velocity is $v_2=\frac{1}{m+M}(p+um+uM).$ This means the kinetic energy of the ball is $\frac{m(p+um+uM)^2}{2(m+M)^2}$ and the kinetic energy of the wall is $\frac{M(p+um+uM)^2}{2(m+M)^2}.$ So the total kinetic energy afterwards is $\frac{(p+um+uM)^2}{2(m+M)}.$

So the change in kinetic energy is $\frac{(p+um+uM)^2}{2(m+M)}-\frac{M(p+um)^2+m(uM)^2}{2mM}=$

$$\frac{mM(p+um)^2+2(p+um)(uM)mM+mM(uM)^2-(mM+M^2)(p+um)^2-(m^2+mM)(uM)^2}{2mM(m+M)}=$$

$$\frac{2(p+um)(uM)mM-M^2(p+um)^2-m^2(uM)^2}{2mM(m+M)}=$$

$$\frac{M\left(2(p+um)(um)-(p+um)^2-(um)^2\right)}{2m(m+M)}=$$

$-\frac{Mp^2}{2m(m+M)}.$ And the minus sign is that I accidentally subtracted initial from final.

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  • $\begingroup$ @Angelika Yes, the change in kinetic energy is the same in both frames. In my work I said that u is speed of the second frame and used conservation of momentum to find the final speeds (an inelastic collision always needs to do this). You take the momentum before and the momentum afterwards and for afterwards you have a common velocity for ball and wall. The total before and after is the same in a fixed frame. Then you can find energies since the kinetic energy of a part is momentum of the part squared over twice the mass of the part. Add for both parts to get total energy. Before & after done $\endgroup$
    – Timaeus
    Sep 3, 2015 at 16:41
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Timaeus has given the full technical answer: the kinetic energy of the wall itself changes a tiny bit. Since kinetic energy scales as $v^2$, this is totally negligible in the first case (where the wall starts with $v = 0$) but actually significant in the second case (where the wall starts with $v = 2$), and that's where the missing energy goes.

Luckily, in general, we know that things will work out when you transform to different reference frames; we've proven it! The nicest way to express this fact is the equation $\partial_\mu T^{\mu \nu} = 0$, where $T$ is the stress-energy tensor; you can expand this out to show that the quantities $\int d^3x T^{0\mu}$ are all conserved. The funny positions of the indices $\mu$ and $\nu$ actually tell us exactly how the quantities $\partial$ and $T$ transform between reference frames, so that just by looking at their positions we know momentum/energy is conserved in any frame.

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Imagine a ball ($mass= 1kg$) moving at a velocity $2 m/s$ towards a wall. When it hits the wall, it suddenly stops, thereby liberating all its KE as heat. Here, $Initial K.E. = (1/2)*m*v^2 = 2J$, and final KE is obviously zero. So heat liberated ($Final KE - Initial KE$) equals $2J$.

Now, suppose I hop on to a car moving at $2 m/s$ towards the oncoming ball from the direction of the wall. So for me, the ball is initially moving at $4 m/s$, and after impact, it is moving at $2 m/s$ (relative velocity). In this case, the $Initial KE=8J$, and the $FinalBallKE=2J$, and the liberated heat is still $2J$. In this case the wall's kinetic energy $WallKE = InitialKE - HeatLiberated - FinalBallKE$, (due to conservation of Energy) equals 4J.

So, your observation was almost correct, just the kinetic energy of the wall was not taken into account.

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    $\begingroup$ When you copied the OP question and made a small change, it makes is VERY difficult to find your answer. $\endgroup$
    – Bill N
    Aug 28, 2015 at 21:23
  • $\begingroup$ @Bill Yes, but the OP may have an unusually easy time understanding what I'm saying - perhaps. Other people - consider the whole thing my answer. The difference is in the final calculation of energies: initial energy becomes heat energy, and kinetic energy of ball, and kinetic energy of wall. $\endgroup$
    – stuffu
    Aug 28, 2015 at 22:13
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In the car frame the ball initially has 8J of kinetic energy relative to you, however it has only 2J of kinetic energy relative to the wall which is also moving in the cars frame.

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  • $\begingroup$ These things are true, but they don't answer the questions about how to reconcile the difference in the resulting $\Delta \mathrm{KE}$. $\endgroup$ Aug 29, 2015 at 0:42

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