Question

A bullet enters into a woden block with speed Vo. its speed becomes half after travelling a distance of 6cm. how much distance will it move before it stops?

Answer 1

Initial speed of bullet $\sf{=v_0.}$

Distance moved after which the speed becomes half $\sf{=6\ cm=0.06\ m.}$

Let the bullet move a distance $\sf{x}$ meters further before it stops. So the total distance covered inside the block will be $\sf{(x+0.06)\ m.}$

Let retardation of the bullet be $\sf{-a.}$

After moving $\sf{0.06\ m}$ distance after entering into the block, speed of bullet becomes halved. So by third equation of motion,

$\longrightarrow\sf{\left(\dfrac{v_0}{2}\right)^2=\left(v_0\right)^2-2a\times0.06}$

$\longrightarrow\sf{\dfrac{(v_0)^2}{4}=\left(v_0\right)^2-0.12a}$

$\longrightarrow\sf{\dfrac{3(v_0)^2}{4}=0.12a}$

$\longrightarrow\sf{(v_0)^2=0.16a\quad\quad\dots(1)}$

After moving $\sf{(x+0.06)\ m}$ distance after entering into the block, speed of bullet becomes zero as it stops. So by third equation of motion,

$\longrightarrow\sf{0^2=\left(v_0\right)^2-2a(x+0.06)}$

$\longrightarrow\sf{\left(v_0\right)^2-2ax-0.12a=0}$

From (1),

$\longrightarrow\sf{0.16a-2ax-0.12a=0}$

$\longrightarrow\sf{0.04a-2ax=0}$

$\longrightarrow\sf{a(0.04-2x)=0}$

Since $\sf{a}$ is non - zero,

$\longrightarrow\sf{0.04-2x=0}$

$\longrightarrow\underline{\underline{\sf{x=0.02\ m}}}$

Or,

$\longrightarrow\underline{\underline{\sf{x=2\ cm}}}$

∴ The bullet moves 2 cm furthermore before it stops. The bullet moves a total of 8 cm inside the block.