Question

A bullet fired into a fixed wooden block loses half of its velocity after penetrating $40\,cm$. It comes to rest after penetrating a further distance of

• A. $\frac{22}{3}cm$
• B. $\frac{40}{3}cm$
• C. $\frac{20}{3}cm$
• D. $\frac{22}{5}cm$

Answer 1

Answer: (B)

$\frac{40}{3}cm$

Answer 2

Let initial velocity of body at point A is v, AB is 40cm.

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