Question

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

Answer 1

Answer: 1.0

Let initial velocity of the bullet $=u$
After penetrating $3cm$ 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 $v^2=u^2-2as$
$\Rightarrow 0=(\frac{u}{2}{)}^2-2(\frac{u^2}{8})\cdot x$
$\Rightarrow x=1cm$.