Answer:
The final velocity of the second player is 6.1 m/s.
Explanation:
The final velocity of the second player can be calculated by conservation of linear momentum (p):
[tex] p_{i} = p_{f} [/tex]
[tex] m_{a}v_{a_{i}} + m_{b}v_{b_{i}} = m_{a}v_{a_{f}} + m_{b}v_{b_{f}} [/tex] (1)
Where:
[tex]m_{a}[/tex]: is the mass of the first football player = 110 kg
[tex]m_{b}[/tex]: is the mass of the second football player = 90 kg
[tex]v_{a_{i}}[/tex]: is the initial velocity of the first football player = 5.0 m/s
[tex]v_{b_{i}}[/tex]: is the initial velocity of the second football player = 0 (he is at rest)
[tex]v_{a_{f}}[/tex]: is the final velocity of the first football player = 0 (he stops after the impact)
[tex]v_{b_{f}}[/tex]: is the final velocity of the second football player =?
By solving equation (1) for [tex]v_{b_{f}}[/tex] we have:
[tex] 110 kg*5.0 m/s + 0 = 0 + 90 kg*v_{b_{f}} [/tex]
[tex] v_{b_{f}} = \frac{110 kg*5.0 m/s}{90 kg} = 6.1 m/s [/tex]
Therefore, the final velocity of the second player is 6.1 m/s.
I hope it helps you!
