A proton A with a speed v moves directly towards a free proton B, originally at rest. Assuming the collision is one dimensional , 

At a beginning the speed of A gradually decreases and the speed B gradually increases. Thus the distance between them decreases. 
When they are in closest approach, their speeds are equal 
Appling law of conservation of linear momentum 
$\begin{array}{l}\mathrm{MV}+0={\mathrm{MV}}^1+{\mathrm{MV}}^1\\ \therefore {\mathrm{V}}^1=\frac{\mathrm{V}}{2}\end{array}$