Answer
$5.65\;\rm m/s$
Work Step by Step
We chose our system to be the man plus the cart and we assume that the cart is rolling without friction which means that this system is isolated. Hence, the momentum of the system is conserved.
$$p_{ix}=p_{fx}$$
$$m_{man}v_{ix,man}+m_{cart}v_{ix,cart}=m_{man}v_{fx,man}+m_{cart}v_{fx,cart}$$
Noting that the two of them initially are having the same initial velocity of 5 m/s.
Thus,
$$(m_{man}+m_{cart})v_{ix}=m_{man}v_{fx,man}+m_{cart}v_{fx,cart}$$
We are given the final velocity of the man relative to the cart, so his final velocity relative to the ground is given by
$$v_{fx,man}=v_{fx,cart}-10$$
Plugging into the previous formula above;
$$ (m_{man}+m_{cart})v_{ix} =m_{man}(v_{fx,cart}-10)+m_{cart}v_{fx,cart}$$
$$ (m_{man}+m_{cart})v_{ix} =m_{man} v_{fx,cart}-10m_{man} +m_{cart}v_{fx,cart}$$
Solving for $v_{fx,cart}$;
$$v_{fx,cart}=\dfrac{ (m_{man}+m_{cart})v_{ix}+10m_{man} }{m_{man}+m_{cart}}$$
Plugging the known;
$$v_{fx,cart}=\dfrac{ 5(70+1000) +(10\times 70) }{70+1000}$$
$$v_{fx,cart}=\color{red}{\bf 5.65}\;\rm m/s$$
