Question

A bullet fired into a fixed target loses half of its velocity after penetrating $3 \,cm$. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion is _____ cm

Answer 1

Let initial velocity of the bullet $ = u$
After penetrating $3 \,cm$ its velocity becomes $\frac{u}{2}$.
From $ v^2 = u^2 - 2as$
$ (\frac{u}{2})^2 = u^2 - 2a(3)$
$\Rightarrow 6a = \frac{3u^2}{4}$
$ \Rightarrow a = \frac{u^2}{8}$
Let it further penetrate through distance $x$ and stops at point $C$. For distance $BC, v = 0, u = u/2, s = x ,a = u^2/8$
From $ v^2 = u^2 - 2as $
$\Rightarrow 0 = ({\frac{u}{2}})^2 - 2(\frac{u^2}{8}) \cdot x$
$\Rightarrow x = 1\,cm$.