Thank you for reporting, we will resolve it shortly
Q.
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
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 $
