Explanation

(b) Let initial velocity of the bullet $= u$
After penetrating $3\ cm$ its velocity becomes $\frac{u}{2}$
[image:3_image186.png]
From $v^{2} = u^{2} - 2as$
$\left( \frac{u}{2} \right)^{2} = u^{2} - 2a(3)$
$\Rightarrow 6a = \frac{3u^{2}}{4} \Rightarrow a = \frac{u^{2}}{8}$
Let further it will penetrate through distance $x$ and stops at point $C$
For distance $BC,\ v = 0,u = u/2,\ s = x,a = u^{2}/8$
For $v^{2} = u^{2} - 2as \Rightarrow 0 = \left( \frac{u}{2} \right)^{2} - 2\left( \frac{u^{2}}{8} \right).x \Rightarrow x = 1\ cm$