Conservation of momentum Steps:

• Define right as positive direction.
• Initial momentum: $80 \times 2.0 + 60 \times (-1.0) = 160 - 60 = 100$ kg·m/s.
• Final momentum of heavier skater: $80 \times 0.8 = 64$ kg·m/s.
• Let $v$ be velocity of lighter skater: $64 + 60v = 100$.
• Solve: $60v = 36$, $v = 0.6$ m/s (speed 0.6 m/s). Why A is correct:
• Conservation of momentum requires the lighter skater's speed after collision to be 0.6 m/s for total momentum to remain 100 kg·m/s. Why the others are wrong:
• B: Assumes lighter rebounds left at 0.6 m/s, giving relative speed $0.8 - (-0.6) = 1.4$ m/s (ignores momentum sign).
• C: Matches initial speed of heavier skater (ignores collision effects).
• D: Approaching speeds added ($2.0 + 0.6$) without using momentum. Final answer: A