A bullet is fired on a target with velocity ' $\mathrm{V}$ '. Its velocity decreases from ' $\mathrm{V}$ '…
A bullet is fired on a target with velocity ' $\mathrm{V}$ '. Its velocity decreases from ' $\mathrm{V}$ ' to ' $\mathrm{V} / 2$ ' when it penetrates $30 \mathrm{~cm}$ in a target. Through what thickness it will penetrate further in the target before coming to rest?
$5 \mathrm{~cm}$
$8 \mathrm{~cm}$
$10 \mathrm{~cm}$
$20\mathrm{~cm}$
Solution
When the velocity of the bullet changes from $\mathrm{V}$ to $\frac{\mathrm{V}}{2}$ the distance travelled by the bullet is $30 \mathrm{~cm}$.
Using $3^{\text {rd }}$ equation of motion,
$\begin{aligned}
& \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{a} \\
& \left(\frac{\mathrm{V}}{2}\right)^2=\mathrm{V}^2+2 \mathrm{a}(30) \\
& \frac{\mathrm{V}^2}{4}=\mathrm{V}^2+60 \mathrm{a} \\
& \frac{-3 \mathrm{~V}^2}{4}=60 \mathrm{a} \\
& \mathrm{a}=\frac{-\mathrm{V}^2}{80}
\end{aligned}$
Further, when a bullet penetrates it comes to rest. So, the final velocity of the bullet becomes zero.
Using the relation,
$\begin{aligned}
& \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as} \\
& 0=\left(\frac{\mathrm{V}}{2}\right)^2+2\left(-\frac{\mathrm{V}^2}{80}\right) \mathrm{s} \\
& \frac{\mathrm{V}^2}{4}=\left(\frac{\mathrm{V}^2}{40}\right) \mathrm{s} \\
& \mathrm{s}=\frac{40}{4} \\
& \mathrm{~s}=10 \mathrm{~cm}
\end{aligned}$