Answer:
The speed of the players after collision = 0 m/s
Step-by-step explanation:
For player A
![\begin{gathered} Mass,\text{ m}_A=80kg \\ Speed,\text{ u}_A=5\text{ m/s} \\ \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/n9hmf50b7pff2zt32lytwt6hb02bvfxx84.png)
For player B
He is moving in opposite direction to player A
![\begin{gathered} Mass,\text{ m}_B=100kg \\ \\ Speed,\text{ u}_B=4\text{ m/s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/su04ewfe7lh81hvhvr9gpyfihequ1iiv4i.png)
Let their speed after collision be v
Using the conservation principle:
![\begin{gathered} m_Au_A+m_Bu_B=(m_A+m_B)v \\ \\ 80(5)+100(-4)=(80+100)v \\ \\ 400-400=180v \\ \\ 0=180v \\ \\ v=(0)/(180) \\ \\ v=0\text{ m/s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/iyzvf45tbv27bbj1fsgof05sb1rul2m3kn.png)
The speed of the players after collision = 0 m/s
This means that the two players come to a halt after collision.