Question

A bullet fired into a fixed wooden block loses half of its velocity after penetrating $40cm$. It comes to rest after penetrating a further distance of

• A. $\frac{40}{3}cm$
• B. $\frac{20}{3}cm$
• C. $\frac{22}{5}cm$
• D. $\frac{26}{5}cm$

Answer 1

The correct option is A

$\frac{40}{3}cm$

Step 1: Given data

Distance, $s=40cm=0.4m$

Final velocity, $v_1$ at distance, $0.4m$is $v_1=\frac{u}{2}$, where, $u$is the initial velocity.

Step 2: Formula used

Equation of motion,

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

Where, $v$is the final velocity, $u$is the initial velocity, $a$ is the deceleration, $s$ is the distance travelled.

Step 3: Find the deceleration $a$ in terms of the final velocity $v$

Substituting, $s=0.4m$

$$\begin{gathered} v_1^2=u^2-2as \\ a=\frac{u^2-{\left(\frac{u}{2}\right)}^2}{2s} \\ a=\frac{4u^2-u^2}{4\times 2s} \\ a=\frac{3u^2}{4\times 2\times 0.4} \\ a=\frac{3u^2}{3.2} \\ a=\frac{15u^2}{16}........\left(i\right) \end{gathered}$$

Step 4: Find the distance travelled by the bullet after it reaches half the initial speed

Using $\left(i\right)$, putting the value of $a$

Let distance after further penetrating $40cm$ is $x$ till it come to rest such that the final velocity, $v_2=0$.

The initial velocity will be, $u^{\text{'}}=\frac{u}{2}$

Again using the equation of motion,

$v^2=u^2-2as$

${v_2}^2=u{\text{'}}^2-2ax$

$x=\frac{({u^{\text{'}}}^2-v_2^2)}{2a}$

$$\begin{gathered} x=\frac{{\left(\frac{u}{2}\right)}^2-0^2}{2\times {\displaystyle \frac{15u^2}{16}}} \\ x=\frac{40}{3}cm \end{gathered}$$

Thus, the distance travelled by the bullet in the wooden block is $\frac{\mathbf{40}}{\mathbf{3}}\mathbf{}\mathit{c}\mathit{m}$.

Hence, option $\left(A\right)$ is the correct option.