Two players, one with mass 75 kg moving at 6 m/s and another with mass 100 kg moving at 8 m/s directly towards each other, collide and stick together. What is their final velocity?

B. 2m/s Step 1: Identify the given values and set up the initial conditions. Let m_1 and v_1 be the mass and initial velocity of the first player, and m_2 and v_2 be the mass and initial velocity of the second player. Given: m_1 = 75 kg v_1 = 6 m/s m_2 = 100 kg v_2 = 8 m/s Since they run directly toward each other, we assign one direction as positive and the other as negative. Let the initial velocity of the first player be positive. v_1 = +6 m/s v_2 = -8 m/s Step 2: Apply the principle of conservation of momentum. In a collision where objects stick together (a perfectly inelastic collision), the total momentum before the collision is equal to the total momentum after the collision. The formula for conservation of momentum is: m_1 v_1 + m_2 v_2 = (m_1 + m_2) V_f where V_f is the final combined velocity of the two players. Step 3: Substitute the values into the conservation of momentum equation. (75 kg)(6 m/s) + (100 kg)(-8 m/s) = (75 kg + 100 kg) V_f 450 kg · m/s - 800 kg · m/s = (175 kg) V_f -350 kg · m/s = (175 kg) V_f Step 4: Solve for the final combined velocity V_f. V_f = -350 kg · m/s175 kg V_f = -2 m/s Step 5: Determine the combined speed. The question asks for the combined speed, which is the magnitude of the final velocity. Speed = |V_f| = |-2 m/s| = 2 m/s The combined speed of the two players just after the collision would be 2 m/s. The final answer is 2m/s.