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?

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Let initial velocity of the bullet = u
After penetrating 3 cm its velocity becomes $\frac{u}{2}$
From $v^2=u^2-2as$
${\left(\frac{u}{2}\right)}^2=u^2-2a\left(3\right)$

$\Rightarrow 6a=\frac{3u^2}{4}\Rightarrow a=\frac{u^2}{8}$
Let further it will penetrate through distance x and stops at point C.
For distance BC, $v=0,u=\frac{u}{2},s=x,a=\frac{u^2}{8}$
From $v^2=u^2-2as\Rightarrow 0={\left(\frac{u}{2}\right)}^2-2\left(\frac{u^2}{8}\right).x\Rightarrow x=1cm.$

Hence the bullet will penetrate 1 cm more.