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# When a bullet is fired at a target, its velocity gets decreased by half after penetrating $30\,cm$ into it. The additional thickness it will penetrate before coming to rest will be:A. $30$B. $10$C. $40$D. $50$

Last updated date: 13th Jul 2024
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Hint:When a bullet is fired and it starts to penetrate in a target, then it will cover some distance with negative acceleration and its final velocity will be zero and we will determine the additional thickness bullet will penetrate by using newton’s equation of motion ${v^2} - {u^2} = 2aS$.

Let us assume in first $30\,cm$ penetration the initial velocity of bullet was $u$ and final velocity behalf of initial velocity as it’s given so we can write as $v = \dfrac{u}{2}$
$S = 0.3\,m$
Now, putting these values in equation ${v^2} - {u^2} = 2aS$ we get,
$\dfrac{{{u^2}}}{4} = {u^2} + 2a(0.3)$
$\Rightarrow \dfrac{{ - 3{u^2}}}{4} = 0.6a \to (i)$

Now, let us assume the bullet further penetrate through distance say $S'$ and final velocity in this case will be zero as bullet will came to rest and initial velocity will be $\dfrac{u}{2}$ , putting these values in equation ${v^2} - {u^2} = 2aS$ we get,
$\dfrac{{ - {u^2}}}{4} = 2aS' \to (ii)$
From equations $(i)and(ii)$ we can write as:
$S' = 0.1\,m$
$\therefore S' = 10\,cm$
So, the thickness to which bullet will further penetrate the target is $S' = 10cm$

Hence, the correct option is B.

Note: It’s important to remember that, the final velocity with which first $30cm$ penetration happen will became the initial velocity of final penetration of $10cm$ and other two equations of motion as $v = u + at$ , $S = ut + \dfrac{1}{2}a{t^2}$ . These three equations are called newton’s equation of motion and whole classical mechanics of motion can be understand through these equations.>