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?

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:
$m_1 = 75 \text{ kg}$
$v_1 = 6 \text{ m/s}$
$m_2 = 100 \text{ kg}$
$v_2 = 8 \text{ m/s}$

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.
$v_1 = +6 \text{ m/s}$
$v_2 = -8 \text{ m/s}$

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.

$$(75 \text{ kg})(6 \text{ m/s}) + (100 \text{ kg})(-8 \text{ m/s}) = (75 \text{ kg} + 100 \text{ kg}) V_f$$ $$450 \text{ kg} \cdot \text{m/s} - 800 \text{ kg} \cdot \text{m/s} = (175 \text{ kg}) V_f$$ $$-350 \text{ kg} \cdot \text{m/s} = (175 \text{ kg}) V_f$$

$$V_f = \frac{-350 \text{ kg} \cdot \text{m/s}}{175 \text{ kg}}$$ $$V_f = -2 \text{ m/s}$$

The question asks for the combined speed, which is the magnitude of the final velocity.

$$\text{Speed} = |V_f| = |-2 \text{ m/s}| = 2 \text{ m/s}$$

The combined speed of the two players just after the collision would be 2 m/s.

The final answer is $\text{2m/s}$.