Is change in kinetic energy invariant?

Question

Consider the inelastic collision between two bodies.

This question follows on from this: Is the coefficient of restitution frame independent and energy conservation?.

One of the answers to this question said that $\Delta E$ had to be in the CofM frame. But is $\Delta E$ not the same in every frame? (I know it is not the same for simply the change in the kinetic energy of one body, but for a collision like this it seems like $\Delta E$ i.e. the energy lost to the soundings should be invariant).

Here is my reasoning consider the frame $S'$ moving at speed $v$ with respect to $S$ (ignoring relativistic effects), then the change in kinetic energy in $S'$ is given by: $$\Delta E'=\frac{1}{2}m_1 u^{'2}_1+\frac{1}{2}m_2 u^{'2}_2-\frac{1}{2}m_1 v^{'2}_1-\frac{1}{2}m_2 v^{'2}_2$$ $$=\frac{1}{2}m_1 (u_1-v)^2+\frac{1}{2}m_2 (u_2-v)^2-\frac{1}{2}m_1 (v_1-v)^2-\frac{1}{2}m_2 (v_2-v)^2$$ $$=(\frac{1}{2}m_1 u^{2}_1+\frac{1}{2}m_2 u^{2}_2-\frac{1}{2}m_1 v^{2}_1-\frac{1}{2}m_2 v^{2}_2)+2v(m_1v_1+m_1v_2-m_1u_1-m_2u_2)+0$$ but due to conservation of momentum the thing in the second brackets is 0 so we are left with: $$\Delta E'=\Delta E$$ Is this correct as it does not feel right? If so can you please explain why it is the case intuitively, thanks.

Answer 1

Let $S$ and $S'$ be the two inertial frame and $S'$ moving with a constant velocity v w.r.t. $S$ frame. Now a force $F$ acting on the particle at point $A$ and displace it to the point $B$. If the position x-coordinates of A point and B point in $S'$ frame are $(x_1',x_2')$ and in $S$ frame are $(x_1,x_2)$ then at any time $t$, $x_1=x_1'+vt$ and $x_2=x_2'+vt$ Since $F=F'$, the work done for displacing the body from point A to B : 1. in $S'$ frame is $F'.(x_2'-x_1')$ 2. in $S$ frame is $ F. (x_2-x_1)=F'.{(x_2'+vt)-(x_1'+vt)}=F'. (x_2'-x_1')$ Hence the work done is same in both frame. So we can say that the change of energy of the body in the both frame is also same.(As you derive) And I think you are right.