The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at $$(4 + x)$$ cm inside the block. The value of $$x$$ is

We are given that a bullet's velocity becomes one-third after penetrating 4 cm in a wooden block, and it faces constant resistance (constant deceleration).

Let the initial velocity of the bullet be $$v_0$$. After penetrating 4 cm, its velocity is $$\frac{v_0}{3}$$. Let the constant deceleration be $$a$$.

Using the kinematic equation $$v^2 = u^2 - 2as$$ for the first 4 cm, $$\left(\frac{v_0}{3}\right)^2 = v_0^2 - 2a(4)$$, which gives $$\frac{v_0^2}{9} = v_0^2 - 8a$$. Solving for $$a$$, we find $$8a = v_0^2 - \frac{v_0^2}{9} = \frac{8v_0^2}{9}$$ and hence $$a = \frac{v_0^2}{9}$$.

When the bullet stops completely, $$v = 0$$, the total stopping distance satisfies $$0 = v_0^2 - 2a \cdot s_{total}$$. Therefore, $$s_{total} = \frac{v_0^2}{2a} = \frac{v_0^2}{2 \cdot \frac{v_0^2}{9}} = \frac{9}{2} = 4.5 \text{ cm}$$.

Since the total distance travelled is $$(4 + x)$$ cm, equating this to 4.5 cm yields $$4 + x = 4.5$$ and hence $$x = 0.5 \text{ cm}$$.

Answer: Option C: 0.5