5 moving points in plane, one goes to infinity

Suppose we have $5$ points in plane, each lying on a line for which no three of these lines intersect in one point, and also non of these $5$ points is an intersection point of two lines. At time $t=0$, each of the points starts to move on its own line in an arbitrary direction and with an arbitrary but constant positive speed. Each point keeps going unless it meets another point. When so, the two points reverse their directions and go back the path they've come. Prove that after some finite time $T$, one of the points will never meet anymore points.

This problem popped into my mind some weeks ago. I kept thinking about it for a long while, but I couldn't reach anything. All the answers are welcomed!

Just in case this helps someone. Let an initial "configuration" be associated with a $n \times n$ matrix $M$ ($n=5$ in the original question), with integer values, with dummy placeholders in the diagonal, and such that each row contains different values, not all negative or positive. Starting from an initial configuration, each iteration does the following:

• decrement all values by one $m_{i,j} := m_{i,j}-1$ ($i\ne j$)
• look for pairs of symmetric zeroes ($m_{i,j}=m_{j,i}=0$) and negate each corresponding row: $m_{i,k} := - m_{i,k}$ $m_{j,k} := - m_{j,k}$

This procedure will either eventually produce some all-negative row ("the point has escaped"), or repeat in a cycle. The conjecture would be that with $n=5$ ( perhaps for all odd $n$?) there cannot be such a cycle.

This model is slightly more general (and hence the impossibility conjecture is stronger) than the original problem ($m_{i,j}$ represents the distance from point $i$ to the line of point $j$, positive if it's approaching it) because it does not take into account the restrictions of having the points along straight lines.