1. **State the problem:** Two ice skaters with masses 54 kg and 73 kg push off each other on a frictionless ice rink. The 54 kg skater moves backward at 3.4 m/sec. We need to find the velocity of the 73 kg skater after the push. 2. **Relevant formula:** Use the law of conservation of momentum, which states that the total momentum before and after the push is the same because no external forces act on the system. $$m_1 v_1 + m_2 v_2 = 0$$ where $m_1 = 54$ kg, $v_1 = -3.4$ m/sec (negative because backward), $m_2 = 73$ kg, and $v_2$ is the unknown velocity of the second skater. 3. **Apply the formula:** $$54 \times (-3.4) + 73 \times v_2 = 0$$ 4. **Solve for $v_2$:** $$73 \times v_2 = -54 \times (-3.4)$$ $$73 \times v_2 = 183.6$$ $$v_2 = \frac{183.6}{73}$$ 5. **Simplify the fraction:** $$v_2 = \frac{\cancel{183.6}}{\cancel{73}} = 2.5137$$ 6. **Final answer:** The velocity of the 73 kg skater after the push is approximately **2.51 m/sec forward**.