Question

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

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

Answer 1

Answer: (B)

$v^2-u^2=2as$
$\Rightarrow {\left(\frac{2u}{3}\right)}^2-u^2=2a{s}_1$
$\Rightarrow \frac{4}{9}{u}^2-u^2=2a{s}_1$
$\Rightarrow \frac{-5u^2}{9\times 2\times s_1}=a$
$\Rightarrow -\frac{5u^2}{18s_1}=a$
Now: $v^2-u^2=2as$
$0-\left(\frac{2u^2}{3}\right)=2\times \left(\frac{5u^2}{18s_1}\right)\times s_2$
$\Rightarrow -\frac{4u^2}{9}=\frac{-2\times 5u^2}{18s_1}\times s_2\Rightarrow \frac{4\times s_1}{5}=s_2$
$\Rightarrow \frac{4\times 4}{5}=s_2\Rightarrow 3.2cm$