Search any question & find its solution
Question:
Answered & Verified by Expert
A bullet of mass $20 \mathrm{~g}$ moving with $500 \mathrm{~ms}^{-1}$ is pierced $1 \mathrm{~cm}$ into a wooden block, then the retarding force experienced by the bullet is
Options:
Solution:
2474 Upvotes
Verified Answer
The correct answer is:
$250 \times 10^3 \mathrm{~N}$
Mass, $\mathrm{m}=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}$
velocity, $\mathrm{u}=500 \mathrm{~m} / \mathrm{s}$
distance, $\mathrm{d}=1 \mathrm{~cm}=0.0 / \mathrm{m}$
final velocity, $v=0$
Using Equation of Motion, $v^2-u^2=2 a d$
$\begin{aligned} & -(500)^2=2 \times \mathrm{a} \times 0.01 \\ & \mathrm{a}=-\frac{500 \times 50000}{2 \times 0.01}=-1.25 \times 10^7 \mathrm{~m} / \mathrm{s}^2\end{aligned}$
using newton's second law,
$\begin{aligned} & \mathrm{F}=\mathrm{ma} \\ & \mathrm{F}=20 \times 10^{-3} \times 1.25 \times 10^7 \\ & \mathrm{~F}=250 \times 10^3 \mathrm{~N}\end{aligned}$
velocity, $\mathrm{u}=500 \mathrm{~m} / \mathrm{s}$
distance, $\mathrm{d}=1 \mathrm{~cm}=0.0 / \mathrm{m}$
final velocity, $v=0$
Using Equation of Motion, $v^2-u^2=2 a d$
$\begin{aligned} & -(500)^2=2 \times \mathrm{a} \times 0.01 \\ & \mathrm{a}=-\frac{500 \times 50000}{2 \times 0.01}=-1.25 \times 10^7 \mathrm{~m} / \mathrm{s}^2\end{aligned}$
using newton's second law,
$\begin{aligned} & \mathrm{F}=\mathrm{ma} \\ & \mathrm{F}=20 \times 10^{-3} \times 1.25 \times 10^7 \\ & \mathrm{~F}=250 \times 10^3 \mathrm{~N}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.