If the bearings are fricionless, the freight car plus robber
has a conserved momentum of zero, meaning from
eqn. 1.37 that the center of mass velocity is zero.
That implies that the center of mass 
is constant.
The robber could be pounding against the wall or running
around in a violent rage,
and this would not alter the position of the center of mass
of the complete system one iota. So if the robber goes
to the left, the freight car must go to the right to compensate.
Let's then make use of center of mass conservation to solve the problem. Let's choose a coordinate system where the center of mass of the freight car is at the origin.
Initially the robber is at a distance 
, and the
mass of robber plus gold is 
.
The center of mass of the system is then
After moving to the door, the freight car has shifted an
amount 
and therefore the center of mass of the
the car is now at . That's what we're interested
in finding. The position of the robber
is now 
. The center of mass is then
Equating the center of mass before with the center of mass after the robber has moved gives:
Moving the first term on the right hand side to the left and factoring the remaining terms of the right hand side gives
or
Plugging in numbers we have
This is greater than 2ft, so the robber would end up as fertilizer for some weeds.
