question_answer

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)         \[\frac{22}{5}cm\]

E)  \[\frac{26}{5}cm\]

Correct Answer: B

Solution :

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           \[x=\frac{40}{3}cm\]