Explanation

(b) Let initial velocity of body a point $A$ is $v,\ AB$ is 40 cm.
From $v^{2} = u^{2} - 2as$
$\Rightarrow \ \ \ \ \ {\ \ \ \left( \frac{v}{2} \right)}^{2} = v^{2} - 2a \times 40$
$Or\ \ \ \ \ \ \ \ \ \ \ \ \ \ a = \frac{3v^{2}}{320}$
Let on penetrating 40 cm in a wooden block, the body moves $x$ distance from $B\ to\ C.$
So, for $B\ to\ C$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u = \frac{v}{2},v = 0$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = x,\ a = \frac{3v^{2}}{320}(deceleration)$
$\therefore\ \ \ \ \ \ \ (0)^{2} = \left( \frac{v}{2} \right)^{2} - 2 \times \frac{3v^{2}}{320} \times x$
$Or\ \ \ \ \ \ \ \ \ \ \ x = \frac{40}{3}cm$