Question

A 98-kg fullback is running along at 8.6 m / s when a 76-kg defensive back running in the same direction at 9.8 m / s jumps on his back. What is the post-collision speed of the two players immediately after the tackle

Answer 1

The total momentum before and after the collision must be conserved.

The total momentum before the collision is:
$p_i = m_1 v_1 + m_2 v_2$
where m1 and m2 are the masses of the two players, and $v_1$ and $v_2$ their initial velocities. Both are considered with positive sign, because the two players are running toward the same direction.

The final momentum is instead
$p_f = (m_1+m_2)v_f$
because now the two players are moving together with a total mass of (m1+m2) and final speed vf.

By requiring that the momentum is conserved
$p_i=p_f$
we  can calculate vf, the post-collision speed:
$m_1 v_1 + m_2 v_2 = (m_1+m_2)v_f$
$v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 +m_2}= \frac{(98 kg)(8.6 m/s)+(76 kg)(9.8m/s)}{98 kg+76 kg}=9.1 m/s$
and the direction is the same as the direction of the players before the collision.