We will have the following:
We solve using the conservation of momentum:
[tex]\begin{gathered} (40kg)(3.5m/s)+(65kg)(0m/s)=(40kg+65kg)v\Rightarrow140kg\ast m/s=(105kg)v \\ \\ \Rightarrow v=\frac{140kg\ast m/s}{105kg}\Rightarrow v=\frac{4}{3}m/s \\ \\ \Rightarrow v\approx1.33m/s \end{gathered}[/tex]So, the final speed of the couple is approximately 1.33 m/s.
