Answer:
6m/s
Explanation:
The concept that we need to give solution to this problem is momentum conservation,
We have here different components or in other words, two different direction for the objects. Then,
[tex]m_1u_1 +m_2u_2 = (m_1+m_2)v_{fx}[/tex]
In this direction is not the man of 75Kg, then we don't considerate his velocity,
[tex]m_1u_1 = (m_1+m_2)v_{fx}[/tex]
Solving for [tex]v_{fx}[/tex]
[tex]v_{fx} = \frac{m_1u_1}{m_1+m_2}[/tex]
[tex]v_{fx} = \frac{(85)(8)}{75+85}[/tex]
[tex]v_{fx} = \frac{(85)(8)}{75+85}[/tex]
[tex]v_{fx} = 4.25m/s[/tex]
In Y direction we have a man, who runs to 9m/s and has a mass of 75kg
In this direction is not the man of 85Kg, then we don't considerate his velocity,
[tex]m_2u_2 = (m_1+m_2)v_{fy}[/tex]
Solving for [tex]v_{fy}[/tex]
[tex]v_{fy} = \frac{m_2u_2}{m_1+m_2}[/tex]
[tex]v_{fy} = \frac{(75)(9)}{75+85}[/tex]
[tex]v_{fy} = \frac{(75)(9)}{75+85}[/tex]
[tex]v_{fy} = 4.21875m/s[/tex]
The speed is like a vector, then the net velocity is
[tex]V= \sqrt{v_{fx}^2+v_{fy}^2}[/tex]
[tex]V = \sqrt{(4.25)^2+(4.21875)^2}[/tex]
[tex]V = 5.988m/s \approx 6m/s[/tex]
