Q110 / 150 · MHT CET 2024 (11 May Shift 1) · Physics
A bullet is fired on a target with velocity V. Its velocity decreases from $v$ to $v / 2$. When it penetrates 30 cm in a target. Through what thickness it will penetrate further in the target before coming to rest?
- A5 cm
- B8 cm
- Correct answerC10 cm
- D12 cm
Solution
When the velocity of the bullet changes from V to $\frac{\mathrm{V}}{2}$ the distance travelled by the bullet is 30 cm . Using $3^{\text {rd }}$ kinematic equation,
$\begin{aligned}
& v^2=u^2+2 a s \\
& \left(\frac{V}{2}\right)^2=V^2+2 a(30) \\
& \frac{V^2}{4}=V^2+60 a \\
& \frac{-3 V^2}{4}=60 a \\
& a=\frac{-V^2}{80}
\end{aligned}$
Further, when a bullet penetrates. it comes to rest. So, the final velocity of the bullet becomes zero. Using the relation, $v^2=u^2+2 \mathrm{as}$ $\begin{aligned} & \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as} \\ & 0=\left(\frac{\mathrm{V}}{2}\right)^2+2\left(-\frac{\mathrm{V}^2}{80}\right) \mathrm{s} \\ & \frac{\mathrm{V}^2}{4}=\left(\frac{\mathrm{V}^2}{40}\right) \mathrm{s} \\ & \mathrm{s}=\frac{40}{4}=10 \mathrm{~cm}\end{aligned}$
Further, when a bullet penetrates. it comes to rest. So, the final velocity of the bullet becomes zero. Using the relation, $v^2=u^2+2 \mathrm{as}$ $\begin{aligned} & \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as} \\ & 0=\left(\frac{\mathrm{V}}{2}\right)^2+2\left(-\frac{\mathrm{V}^2}{80}\right) \mathrm{s} \\ & \frac{\mathrm{V}^2}{4}=\left(\frac{\mathrm{V}^2}{40}\right) \mathrm{s} \\ & \mathrm{s}=\frac{40}{4}=10 \mathrm{~cm}\end{aligned}$
Hence, the correct answer is option: 'C'.