Thermal energy generated due to loss in kinetic energy when observed from two different frames of reference A body is moving with a velocity $v$ with respect to a frame of reference $S_1$.
It bumps into a very heavy object and comes to rest instantaneously, its kinetic energy 
$$\frac{1}{2}mv^2$$ 
as seen from the frame $S_1$ is completely converted to thermal energy.
Now, a man moving with a uniform velocity $V$ (in the direction of the body) with respect to $S_1$ observes the body. He notes that its initial kinetic energy of the body is 
$$\frac{1}{2}m(v+V)^2,$$ 
and  that after it rams into the heavy body as 
$$\frac{1}{2}mV^2,$$ 
and concludes that the thermal energy produced is the difference 
$$\frac{1}{2}m(v+V)^2-\frac{1}{2}mV^2~=~\frac{1}{2} mv^2+mvV.$$     
Which of the two answers is correct and why?  
 A: The first answer is correct.  The second one is wrong because it ignores the energy of the heavy body.  
In your example, imagine some putty hitting a wall.  The putty is moving to the right and the wall is stationary.  When there's a collision, the wall picks up some very tiny velocity $w$.  That velocity is basically inversely-proportional to the wall's mass $M$.  The kinetic energy it picks up is proportional to $Mw^2$, and thus also inversely proportional to $M$.  Therefore, in this frame the wall's energy is negligible.
However, if there is a man moving to the left, he sees the putty moving faster, as you say, but he also sees the wall moving to the right.  When the putty hits it, the wall will move a tiny bit faster to the right and pick up a little bit more kinetic energy.  This time, though, the wall's energy gain cannot be made negligible by making the wall more massive.  You can work out easily that in the limit $M\to\infty$ the kinetic energy the wall gains is $mvV$, compensating for that lost by the putty in this frame.
Here is a general solution.  The kinetic energy of a particle is
$$T = \frac{\mathbf{p}^2}{2m}$$
If we boost into a frame moving at speed $\mathbf{v}$ relative to the original, the new momentum is
$$\overline{\mathbf{p}} = \mathbf{p} + m\mathbf{v}$$
the new kinetic energy works out to be 
$$\overline{T} = T + \mathbf{p}\cdot\mathbf{v} + \frac{m\mathbf{v}^2}{2}$$
Thus the kinetic energy "added" by viewing the process in a new frame is
$$\Delta T = \mathbf{p}\cdot\mathbf{v} + \frac{m\mathbf{v}^2}{2}$$
If you have two particles, this becomes
$$\Delta T = (\mathbf{p_1 + \mathbf{p}_2})\cdot\mathbf{v} + \frac{(m_1 + m_2)\mathbf{v}^2}{2}$$
which is exactly the same as the expression for a single particle of momentum $\mathbf{p}_1 + \mathbf{p}_2$ and mass $m_1 + m_2$.
Say you are watching two particles.  The energy is $E_1$.  Before they collide, you boost to a new frame.  In that new frame, the new kinetic energy is 100J greater, so the energy is $E_1 + 100J$.  Then the particles collide.  The energy becomes $E_2$.
You could also have waited for the particles to collide first, then boosted to a new frame.  In that case, the energy after the collision and before the boost is $E_f$.  What we showed above is that the boost still results in a gain of 100J, so the energy after the boost is $E_f + 100J$.  
It is evident that $E_2 = E_f + 100J$ because they are the same physical situation.  Therefore
$$E_2 - (E_1 + 100J) = E_f + 100J - (E_1 + 100J) = E_f - E_1$$
The left hand side is the energy lost during the collision in the boosted frame.  The right hand side is the energy lost during the collision in the original frame.  Thus, when there is an inelastic collision going on, boosting to a new frame only changes the total amount of kinetic energy around.  It does not change the amount of kinetic energy transformed to thermal energy.
A: A good start to understand conservation of energy.
Energy is not conserved because you have not taken into conservation of momentum. When it bumps into a heavy object that heavy object moved. But you did not take into account that part. I guess conservation of momentum is more fundamental than conservation of energy. Do the above equation once again with conservation of momentum taken into account.
