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

$\frac{40}{3}cm$

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B

$\frac{20}{3}cm$

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C

$\frac{22}{5}cm$

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D

$\frac{26}{5}cm$

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Solution

## The correct option is A $\frac{40}{3}cm$Step 1: Given dataDistance, $s=40cm=0.4m$Final velocity, ${v}_{1}$ at distance, $0.4m$is ${v}_{1}=\frac{u}{2}$, where, $u$is the initial velocity.Step 2: Formula usedEquation 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, $s=0.4m$ ${v}_{1}^{2}={u}^{2}-2as\phantom{\rule{0ex}{0ex}}a=\frac{{u}^{2}-{\left(\frac{u}{2}\right)}^{2}}{2s}\phantom{\rule{0ex}{0ex}}a=\frac{4{u}^{2}-{u}^{2}}{4×2s}\phantom{\rule{0ex}{0ex}}a=\frac{3{u}^{2}}{4×2×0.4}\phantom{\rule{0ex}{0ex}}a=\frac{3{u}^{2}}{3.2}\phantom{\rule{0ex}{0ex}}a=\frac{15{u}^{2}}{16}........\left(i\right)$Step 4: Find the distance travelled by the bullet after it reaches half the initial speedUsing $\left(i\right)$, 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$ $x=\frac{\left({{u}^{\text{'}}}^{2}-{v}_{2}^{2}\right)}{2a}$ $x=\frac{{\left(\frac{u}{2}\right)}^{2}-{0}^{2}}{2×\frac{15{u}^{2}}{16}}\phantom{\rule{0ex}{0ex}}x=\frac{40}{3}cm$Thus, the distance travelled by the bullet in the wooden block is $\frac{\mathbf{40}}{\mathbf{3}}\mathbf{}\mathbit{c}\mathbit{m}$.Hence, option $\left(A\right)$ is the correct option.

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