Force of a Train Imagine that there are two trains and  the first train is twice as long as the second train. They have the same mass per unit length and they are traveling at exactly the same speed.
If the first train hit me, would it hit me with twice as much force as the second train?  These are two distinct situations: 1) I am hit by the first train only, 2) I am hit by the second train only.
Force is mass times acceleration, so if the one train has twice the mass, then it seems likely that it would have twice the force.  But I am not sure.
 A: First of all you should note that Newton's law says when $F$ acts on a mass $m$, then that mass will move with acceleration $a$.
Here, we should apply the laws of collision and by using the conservation of momentum, find out what your velocity will be after the collision. Before collision we have: $p_{tot}=mv$ and after collision $p_{tot}'=mv'+MV$ where $M$ and $V$ are your mass and speed, $m$ and $v$ are mass and speed of train(s). Also for energy, we have $$mv^2=mv'^2+MV^2$$Now by putting $p_{tot}=p_{tot}'$ and solving the equations, we find $$V=\frac{2mv}{M+m}.$$
Now you can see the bigger train will give you more speed (or more momentum) and so the collision is harder, which means the change of your momentum is more. (recall $F=\frac{dp}{dt}$). On the other hand, since $M \ll m$, so this exceeding is not obvious and maybe we can say the effect of both trains are similar.
A: The difference in force to stop the trains you are talking about here is the difference in force is needed to bring the train to a stop within a particular distance Let me tell you what I mean.
When you try to stop the train, you'll obviously be dragged in front of the train. Say the dragging causes an uniform force (due to the friction from the ground) $f_1$ to act on the heavier train, and $f_2$ to act on the other. $f_1$ and $f_2$ act such that both trains come to rest within the same distance $x$. You say that both trains travel with the same velocity $v$, but have different masses. So, let the heavier train have mass $M$ and the lighter one have mass $m$. 
The trains thus have kinetic energies $\frac 12 M v^2$ and $\frac 12 m v^2$. 
In stopping the trains, the work you'll have to do to is $W_1=f_1x=\frac 12 Mv^2$ for the heavier train and $W_2=f_2x=\frac 12 mv^2$ for the other one. Once we divide the two, $${f_1x \over f_2x}={\frac 12 Mv^2 \over \frac 12 mv^2}\implies \frac {f_1}{f_2}=\frac {M}{m}$$ 
Thus, $$f_1 > f_2$$
By Newton's third law, the force you exert on the train is equal what the train exerts on you. Thus $$\boxed {force \ exerted \ by \ the \ heavier \ train \ on \ you \ > \ force \ exerted \ by \ the \ lighter \ train \ on \ you}$$
