A bullet is fired normally on an immovable wooden plank. It loses $25\%$ of its momentum in penetrating a thickness of $3.5 cm$. The total thickness penetrated by the bullet is

Let $u cm{s}^{-1}$ be the speed of the bullet. Since the mass of the bullet remains unchanged, its speed becomes $v=\frac{3u}{4}cm{s}^{-1}$ after it penetrates a distance $x=3.5cm.$

The retardation a due to the resistance of the wooden is given by 

$$\begin{gathered} u^2-v^2=2ax \\ or u^2-{\left(\frac{3u}{4}\right)}^2=2a\times 3.5 \end{gathered}$$

which gives $a=\frac{u^2}{16}cm{s}^{-2}.$

The bullet will come to rest when its velocity$v\text{'} = 0$. If $x\text{'}$is the thickness penetrated by the bullet, then 

$$\begin{gathered} u^2-v{\text{'}}^2=2ax\text{'} \\ or x\text{'}=\frac{u^2}{2a}. \\ But a=\frac{u^2}{16}cm{s}^{-2}. \end{gathered}$$

Therefore, $x\text{'}=\frac{u^2\times 16}{2u^2}=8cm$