The initial kinetic energy, ${\left(K.E.\right)}_i=\frac{1}{2}m{v}^2$, and final kinetic energy, 
${\left(K.E.\right)}_f=\frac{1}{2}\frac{m}{2}v{\text{'}}^2$, given,
 $$\begin{gathered} \frac{1}{2}\times \frac{m}{2}v{\text{'}}^2=\frac{1}{2}\times m{v}^2\times 16 \\ \Rightarrow v\text{'}=\sqrt{32}v \end{gathered}$$

then initial momentum, $mv$ and final momentum, $\frac{m}{2}v\text{'}=\frac{m}{2}4\sqrt{2}v=2\sqrt{2}mv$, 

so net momentum is increased by $2\sqrt{2}$, times.