Two-dimensional collision conservation of energy

Question

According to wikipedia (https://en.wikipedia.org/wiki/Elastic_collision), and the sources cited in the article, the new velocity after a two-dimensional, elastic collision, can be calculated by rotating the reference frame so that components perpendicular to the line of contact do not change, and the use the formulas for a one-dimensional elastic collision for the remaining components along the line of contact. But that does not seem to be right to me, because these equations are derived using both conservation of momentum and the conservation of energy. And while it seems perfectly fine to apply the conservation of momentum to this "reduced, one-dimensional problem", I don't see why it should be possible to "split" the conservation of energy into parts. Can anyone explain to me why that's possible?

Answer 1

We can write the kinetic energy of a particle moving in 2 dimensions as $$E=\frac{1}{2}mv^2=\frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2$$ Where $v_x$ and $v_y$ are the perpendicular components of velocity. If we take $v_x$ to be the velocity perpendicular to normal, the conservation of energy for $2$ colliding particles $1$ and $2$ becomes:

$$\frac{1}{2}m_1(u_{1x}^2 +u_{1y}^2) +\frac{1}{2}m_2(u_{2x}^2 +u_{2y}^2)=\frac{1}{2}m_1(u_{1x}^2 +v_{1y}^2)+\frac{1}{2}m_2(u_{2x}^2 +v_{2y}^2)$$

Where I used the fact $v_{1x}=u_{1x}$ and $v_{2x}=u_{2x}$. Thus, after cancellation, we get:

$$\frac{1}{2}m_1u_{1y}^2 +\frac{1}{2}m_2u_{2y}^2=\frac{1}{2}m_1v_{1y}^2+\frac{1}{2}m_2(u_{2y}^2)$$

Which is the same as that of the $1$ dimensional case.