Unfortunately, that StackExchange discussion only deals with elastic collisions. An arrow hitting wood is an inelastic collision. But the theory still works if we consider conservation of momentum, and that the change in kinetic energy is what matters, before and after, rather than the absolute value of kinetic energy.
For example, let's suppose two lumps of dough, each weighing 1 kg, collide, and after the collision they stick together. Not sure how or why this might happen, but whatever. To an outside observer, each lump of dough is travelling at 5 m/s.
Case 1: From the outside observer's frame of reference, each lump of dough has a kinetic energy of 0.5 x (1 kg) x (5 m/s)² = 12.5 J, for a total of 25 J. After the collision, the combined lump of dough is stationary, so it has a kinetic energy of 0 J. The change in kinetic energy during the collision was 25 J − 0 J = 50 J.
Case 2: From the perspective of a fly sitting on one dough ball, it is stationary, but the other dough ball is approaching at -10 m/s. So the total kinetic energy before the collision is 0 J plus 0.5 x (1 kg) x (-10 m/s)² = 50 J. Assuming the fly didn't get squashed, it will find the combined lump of dough after the collision is now moving backwards* at -5 m/s. So the new kinetic energy is 0.5 x (2 kg) x (-5 m/s)² = 25 J. Therefore the change in kinetic energy is 50 J − 25 J = 25 J. The change in kinetic energy is 25 J, the same as before.
* Why it's moving backwards at 5 m/s is conservation of momentum: We have (1 kg) x (0 m/s) + (1 kg) x (-10 m/s) = (2 kg) x (? m/s), therefore the new velocity = -5 m/s to balance the equation.