Topher925 said:
The mass is constant, but the center of mass IS moving.
No, it is not. Assume we have a system of
N constant mass particles that interact with one another but not with anything else and do so in concordance with Newton's laws of motion. In other words, the only forces acting on the particles are internal; there are no external forces. The linear momentum of the system as a whole is the sum of the linear momentum of each particle in the system:
\mathbf p_{\text{sys}} = \sum_{i=1}^N m_i \frac {d \mathbf r_i}{dt}
Differentiating with respect to time,
\frac{d\mathbf p_{\text{sys}}}{dt} =<br />
\sum_{i=1}^N m_i \frac {d^2 \mathbf r_i}{dt^2}
By Newton's second law, the force acting on the
ith particle is
m_i\frac {d^2 \mathbf r_i}{dt^2} = \sum_{j=1,j\ne i}^N\mathbf F_{ij}
where \mathbf F_{ij} is the force on the
ith particle due to the
jth particle. A particle applies zero net force on itself, or \mathbf F_{ii} = 0. With this, the above becomes
m_i\frac {d^2 \mathbf r_i}{dt^2} = \sum_{j=1}^N\mathbf F_{ij}
Applying this to the time derivative of the system momentum vector,
\frac{d\mathbf p_{\text{sys}}}{dt} =<br />
\sum_{i=1}^N \sum_{j=1}^N \mathbf F_{ij}
Rewriting the sum by pairing the \mathbf F_{ij} and \mathbf F_{ji} terms,
\frac{d\mathbf p_{\text{sys}}}{dt} =<br />
\sum_{i=1}^N \sum_{j=1}^i \mathbf F_{ij} + \mathbf F_{ji}
By Newton's third law, \mathbf F_{ji} =- \mathbf F_{ij}, and thus
\frac{d\mathbf p_{\text{sys}}}{dt} = 0
In other words, the momentum of a system of particles that is not subject to any external forces is conserved regardless of any internal interactions amongst the particles so long as those internal interactions obey Newton's third law.
In the problem at hand, the initial velocities of the man and board are both zero and the initial total momentum is thus zero. Since no external forces act on the man-board system, the total momentum is constant: zero. The combined center of mass does not move.
Woohoo! 
