Newton's Law in space Assume a spaceship that is at rest w.r.t. an external observer. For simplicity sake assume it is a simple hollow cylinder (length >> diameter). Inside a spaceship there is a person who runs from one end of cylinder, gains some kinetic energy inside the ship and hits the opposite wall of the cylinder. 
What would happen to the cylinder w.r.t. to the external observer. Intuition says the ship should gain some kinetic energy. Is that correct?
 A: Let the ship be orientated along the $\hat i$ direction and the person and the cylinder start from rest relative to an external observer..  
Initial momentum $= 0 \hat i$
Final momentum = $mu\,\hat i+ MU \hat i$ where $m$ and $M$ are the masses of the person and the tube and $u$ and $U$ are their components of velocity as measured by an external observer, who was initially at rest relative to the cylinder, in the $\hat i$ direction.
So using the conservation of linear momentum, $mu = - MU$ and the total kinetic energy is $\frac 12 mu^2 + \frac 12 MU^2$.  
Now it all depends as to what actually happens when the person hits the wall.  
If the person sticks to the wall the final result is that the final velocity of the cylinder and the wall is zero (conservation of linear momentum of the person-cylinder system as there are no external forces acting on the system ie centre of mass of the system does not move) and the final kinetic energy of the person and the cylinder is zero.
The collision between the person and the wall must have been inelastic and the kinetic energy of the cylinder and the person before the person hit the wall must have been converted into heat and done work to permanently deform the person and/or the wall ie break bonds.
All that energy coming from the work done by the person using chemical energy. 
The other extreme is that there is an elastic collision between the person and the wall where kinetic energy of the cylinder-person system is conserved.  
Now one would have to solve for the velocities of the person and the cylinder after rebound using the conservation of linear momentum and the conservation of kinetic energy.
$0 \hat i = mv \hat i+ MV \hat i$ and $\frac 12 mu^2 + \frac 12 MU^2 = \frac 12 mv^2 + \frac 12 MV^2$ where $v$ and $V$ are the components of the velocity of the person and the cylinder in the $\hat i$ direction after rebounding from the wall.
So the total linear momentum of the person-cylinder system would still be zero (centre of mass not moving) but the system would have kinetic energy.
In between the person sticking to the wall and elastically rebounding off the wall there are a who range of possible options but in each of those cases the person and the cylinder would have kinetic energy after the person hit the wall but the centre of mass of the person-cylinder system would no move.
A: No, it is not correct.  Conservation of linear momentum tells you, immediately, that unless the spaceship ejects some mass it can not change its velocity.
Note, however, that while the person is running along  inside the ship it will move (and if, suddenly, the wall at the end of the ship was not there and the running person fell from the ship then it would continue to move, as it now has ejected some mass).  However the centre of mass of the combined ship-person system will not move in either case (by 'move' here I mean change velocity with respect to some inertial frame, of course).
A: I believe you are correct. Using conservation of linear momentum:
$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$
Where the mass of the person is $m_1$, his speed before he hits the wall is $u_1$, the mass of the cylindrical space ship is $m_2$ and the initial velocity of the cylinder is $u_2$. $v_1$ and $v_2$ are their respective velocities post collision. 
The cylinder is at rest so $u_2 = 0$. Assuming that the person running comes to rest once he hits the wall:
$$m_1u_1 = m_2v_2$$
$$v_2 = \frac{m_1u_1}{m_2}$$
Therefore, $v_2 > 0$, so the cylinder now has some kinetic energy. It will be a very small amount of kinetic energy, because $m_2 >> m_1u_1$, but it will have kinetic energy nonetheless. 
If the person were to rebound, $v_1$ would be in the negative direction, so the equation would look like this:
$$m_1u_1 = m_2v_2 - m_1v_1$$
$$v_2 = \frac{m_1(u_1 + v_1)}{m_2}$$
In this case, $v_2$ would be even greater.
So yes, in order to conserve momentum, the ship would have to have some kinetic energy, no matter how small, post collision.
