07:13
I'll have to work through a few examples to see what's going on ...
Suppose the initial velocity is u = (ux, uy) and the final velocity is v = (vx, vy)
Then initial KE is ¹⁄₂mu² = ¹⁄₂mux² + ¹⁄₂muy²
And likewise final KE is ¹⁄₂mux² + ¹⁄₂mvy²
So ΔKE = (¹⁄₂mvx² + ¹⁄₂mvy²) - (¹⁄₂mux² + ¹⁄₂muy²)
= (¹⁄₂mvx² - ¹⁄₂mux²) + (¹⁄₂mvy² - ¹⁄₂muy²)
> And likewise final KE is ¹⁄₂mux² + ¹⁄₂mvy²
Typo: And likewise final KE is ¹⁄₂mvx² + ¹⁄₂mvy²