Answer:
[tex]v_2=-8.33m/s[/tex]
Explanation:
The linear momentum during this collision must be conserved, which means that momentum before and after the collision must be the same.
We have a big defensive lineman of mass [tex]m_1=150Kg[/tex] and velocity [tex]v_1=5m/s[/tex] going in what we will call the positive direction, and a wide receiver of mass [tex]m_2=90Kg[/tex] and velocity [tex]v_2[/tex] which we don't know. Since the collision is head on, momentum before the collision will be [tex]p_i=m_1v_1+m_2v_2[/tex]
After the collision, the receiver drops the lineman in his tracks, which means they come to a stop, so the momentum is null, and since this momentum must be equal to the one before the collision we have:
[tex]p_i=0=m_1v_1+m_2v_2[/tex]
Which means:
[tex]m_1v_1=-m_2v_2[/tex]
So for [tex]v_2[/tex] we have:
[tex]v_2=-\frac{m_1v_1}{m_2}=-\frac{(150Kg)(5m/s)}{(90Kg)}=-8.33m/s[/tex]
where the negative sign indicates its opposite to our positive direction (of the big defensive lineman).
