The velocity of the bullet becomes one third after it penetrates 4   cm in a wooden block. Assuming…

According to the given information for the first $4 \mathrm{cm}$, $u=v_0$ and $v=\frac{v_0}{3}$. Applying equation of motion for constant acceleration,

$v^2=u^2+2as$

$$\begin{gathered} \Rightarrow {\left(\frac{v_0}{3}\right)}^2=v^2+2a\left(4\right) \\ \Rightarrow a=-\frac{8 {v_0}^2}{9\left(8\right)}=-\frac{{v_0}^2}{9} \end{gathered}$$

When the body stops, $v=0$

$\Rightarrow 0={v_0}^2+2a\left(4+x\right)$

$\Rightarrow {v_0}^2=2\left(\frac{{v_0}^2}{9}\right)\left(4+x\right)$

$\Rightarrow 4.5=4+x$

$\Rightarrow \mathrm{x}=0.5 \mathrm{cm}$