1. **State the problem:** We have two football players colliding inelastically and coming to a complete stop after the collision. Player A has mass $m_A = 75$ kg and velocity $v_A = +5$ m/s. Player B has mass $m_B = 60$ kg and an unknown velocity $v_B$ moving in the opposite direction. We need to find $v_B$ such that the final velocity after collision is zero. 2. **Relevant formula:** In an inelastic collision where two objects stick together, momentum is conserved. The total momentum before collision equals the total momentum after collision. $$m_A v_A + m_B v_B = (m_A + m_B) v_f$$ Since they come to a complete stop after collision, $v_f = 0$. 3. **Apply the formula:** $$75 \times 5 + 60 \times v_B = (75 + 60) \times 0$$ Simplify: $$375 + 60 v_B = 0$$ 4. **Solve for $v_B$:** $$60 v_B = -375$$ $$v_B = \frac{-375}{60} = -6.25$$ 5. **Interpretation:** Player B must be moving at $-6.25$ m/s (opposite direction to Player A) before the collision for both to come to rest after colliding inelastically. **Final answer:** $$v_B = -6.25 \text{ m/s}$$