Relative speed conservation in elastic collisions Steps:

• Assign directions: ball X has initial velocity +u_x, ball Y has -u_y (speeds u_x, u_y positive).
• Relative velocity of approach: u_x - (-u_y) = u_x + u_y.
• For perfectly elastic collision, relative velocity reverses: post-collision relative velocity = -(u_x + u_y).
• After collision, velocities -v_x and +v_y (speeds v_x, v_y positive), so relative velocity v_y - (-v_x) = v_x + v_y.
• Thus, u_x + u_y = v_x + v_y.

Why A is correct:

• Matches the coefficient of restitution e=1, where speed of separation equals speed of approach: u_x + u_y = v_x + v_y.

Why the others are wrong:

• B: Resembles momentum conservation for equal masses but ignores signs and relative velocity reversal.
• C: Incorrect algebraic rearrangement; does not follow from conservation laws.
• D: Duplicate of B; same error in applying momentum without masses specified.

Final answer: A