Kinetic energy
Work-energy theorem
Procedure: decomposition
one case
Requirement: The speed in one direction is unchanged Explanation: reference frame conversion
Proof for one case: \( \frac{1}{2} m v_1^2 - \frac{1}{2} m v_0^2 = \frac{1}{2} m (\sqrt{v_0^2 + v_y^2})^2 - \frac{1}{2} m v_0^2 = \frac{1}{2} m v_y^2 \)
general
\( F_x = m \frac{v_x^2 - v_{x0}^2}{2x} \)
\( W_x = F_x x = \frac{1}{2}m v_x^2 - \frac{1}{2}m v_{x0}^2\)