Question

If a bullet losses one-third of its velocity on penetrating $4 \,cm$ in a wooden block, then how much will it penetrate more before coming to rest?

• A. $1\, cm$
• B. $3.2 \,cm$
• C. $3 \,cm$
• D. $2.7 \,cm$

Answer 1

Answer: (B)

$v^{2}-u^{2}=2 a s$
$\Rightarrow \left(\frac{2 u}{3}\right)^{2}-u^{2}=2 a s_{1}$
$\Rightarrow \frac{4}{9} u^{2}-u^{2}=2 a s_{1}$
$\Rightarrow \frac{-5 u^{2}}{9 \times 2 \times s_{1}}=a$
$\Rightarrow -\frac{5 u^{2}}{18 s_{1}}=a$
Now: $v^{2}-u^{2}=2 a s$
$0-\left(\frac{2 u^{2}}{3}\right)=2 \times\left(\frac{5 u^{2}}{18 s_{1}}\right) \times s_{2}$
$\Rightarrow -\frac{4 u^{2}}{9}=\frac{-2 \times 5 u^{2}}{18 s_{1}} \times s_{2} \Rightarrow \frac{4 \times s_{1}}{5}=s_{2}$
$\Rightarrow \frac{4 \times 4}{5}=s_{2} \Rightarrow 3.2\, cm$