# Thermal energy generated by collision observed from two different frames of reference

An isolated system is composed of two bodies $A$ and $B$, with masses $m_A$ and $m_B$, $m_A \ne m_B$, which are in route of collision. The relative velocity between them is $v$. The collision is inelastic and I want to calculate how much thermal energy is generated at the collision.

I understand that the thermal energy $\Delta T$ generated equals the variation of the kinetic energy, so that the total energy is conserved. I expected $\Delta T$ to be independent of the reference frame.

However, if I consider a reference frame $S_A$ fixed in body $A$, I get the following change in kinetic energy of the system:

$$\Delta T_A = \left( \frac{m_A 0^2}{2} + \frac{m_B v^2}{2} \right) - \left( \frac{m_A 0^2}{2} + \frac{m_B 0^2}{2} \right) = \frac{m_B v^2}{2}.$$

And if I consider a reference frame $S_B$ fixed in body $B$, I get a different change in the kinetic energy of the system:

$$\Delta T_B = \left( \frac{m_A v^2}{2} + \frac{m_B 0^2}{2} \right) - \left( \frac{m_A 0^2}{2} + \frac{m_B 0^2}{2} \right) = \frac{m_A v^2}{2}.$$

Could you please point out the flaw in this reasoning?

• The final velocity cannot be $0$ in either of these frames. Momentum is conserved, even in inelastic collisions.
– Mike
Feb 16 '18 at 0:56
• If after the collision both bodies stick together, and if the reference frame is fixed in one of them, the final velocity is 0 with respect to the reference frame, isn't it? Feb 16 '18 at 0:59
• It would be interesting to have a similar question addressing the conservation of the momentum, and why this reasoning would lead to the false conclusion that momentum is not conserved. Feb 16 '18 at 1:01
• One more minor comment: Usually the change in a quantity is defined as the final value minus the initial value — rather than initial minus final as you've written here (and as I've kept to in my answer). It makes a bit more sense to get a negative result when the kinetic energy is decreasing.
– Mike
Feb 16 '18 at 2:34

So a correct analysis of your problem could use a frame $F_A$ that starts out with the same velocity as body $A$, or a frame $F_B$ that starts out with the same velocity as body $B$... or literally any inertial frame, though it's probably easiest in either of those two. Using conservation of momentum, and arbitrarily picking the sign of $v$, we have \begin{align} F_A: \qquad -m_B\, v &= (m_A+m_B)v_{f, A} &\qquad \Delta T_A &= \frac{m_B v^2}{2} - \frac{(m_A+m_B)v_{f,A}^2}{2} \\ F_B: \qquad \hphantom{-}m_A\, v &= (m_A+m_B)v_{f, B} &\qquad \Delta T_B &= \frac{m_A v^2}{2} - \frac{(m_A+m_B)v_{f,B}^2}{2}. \end{align} Solve either of the equations on the left for either $v_f$, plug that into the corresponding equation on the right, and you get $$\Delta T_A = \Delta T_B = \frac{m_A m_B}{m_A+m_B} \frac{v^2}{2}.$$ It's the same for either frame.
Another good frame is the center-of-mass frame, $F_{AB}$, in which the velocity of the final combined object is $0$. Momentum conservation is expressed by $$m_A v_A + m_B v_B = 0,$$ and by the definition of the relative velocity $v$ we have $$v_A - v_B = v.$$ Various rearrangements of these equations let us write \begin{align} v_A &= \frac{m_B}{m_A+m_B} v, \\ v_B &= -\frac{m_A}{m_A+m_B} v. \end{align} So we get $$\Delta T_{AB} = \left( \frac{m_A v_A^2}{2} + \frac{m_B v_B^2}{2} \right) - \left( \frac{m_A 0^2}{2} + \frac{m_B 0^2}{2} \right) = \frac{m_A m_B}{m_A+m_B} \frac{v^2}{2}.$$ Again, it's the same answer.
The problem is that you assume that the reference frames $S_A$ and $S_B$ stay the inertial frames before and after the collision even though their velocities change abruptly at the collison. It would be better to use the center of mass reference frame which does not change with the collision.