Assumption: The wooden block offers a constant resistive force, so the retardation is constant.

Let the body enter the block with speed $u$.

After penetrating $s_1 = 3\text{ cm}$, its speed becomes half: $v = \frac{u}{2}$

Use the kinematic relation: $v^2 = u^2 + 2as$

For the first $3\text{ cm}$: $\left(\frac{u}{2}\right)^2 = u^2 + 2a(3)$ $\frac{u^2}{4} = u^2 + 6a$ $6a = -\frac{3u^2}{4}$ $a = -\frac{u^2}{8}$ (with distance in cm units)

Now let the additional distance to stop be $s_2$.

Initial speed for this part is $u/2$, final speed is $0$.

Again, $0 = \left(\frac{u}{2}\right)^2 + 2as_2$ $0 = \frac{u^2}{4} + 2\left(-\frac{u^2}{8}\right)s_2$ $0 = \frac{u^2}{4} - \frac{u^2}{4}s_2$

Dividing by $u^2/4$: $0 = 1 - s_2$ $s_2 = 1\text{ cm}$

Answer: The body penetrates $1\text{ cm}$ more before coming to rest.

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