1. **State the problem:** Two football players collide inelastically and become entangled. Player A has weight 686 N and velocity +6 m/s, Player B has weight 784 N and velocity -7 m/s. Find their combined velocity after collision. 2. **Relevant formula:** In an inelastic collision with no external forces, momentum is conserved. Total momentum before collision equals total momentum after collision. $$m_A v_A + m_B v_B = (m_A + m_B) v_f$$ where $m_A$, $m_B$ are masses, $v_A$, $v_B$ are initial velocities, and $v_f$ is final velocity. 3. **Convert weights to masses:** Weight $W = mg$, so mass $m = \frac{W}{g}$ with $g = 9.8$ m/s². $$m_A = \frac{686}{9.8} = 70\, \text{kg}$$ $$m_B = \frac{784}{9.8} = 80\, \text{kg}$$ 4. **Calculate total momentum before collision:** $$p_{total} = m_A v_A + m_B v_B = 70 \times 6 + 80 \times (-7) = 420 - 560 = -140\, \text{kg m/s}$$ 5. **Calculate combined mass:** $$m_{total} = 70 + 80 = 150\, \text{kg}$$ 6. **Find final velocity $v_f$:** $$v_f = \frac{p_{total}}{m_{total}} = \frac{-140}{150} = -0.9333\, \text{m/s}$$ 7. **Interpretation:** The negative sign means the combined players move in the direction of Player B's initial velocity after collision. **Final answer:** $$v_f = -0.93\, \text{m/s}$$