Ask Questions, Get Answers

Want to ask us a question? Click here
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11
0 votes

A bullet fired into a plank loses half of its velocity after penetrating a thickness of $3 cm$. assuming that it has constant force of resistance, how much further does it penetrate before coming to rest

\[(a)\;4\;cm\quad (b)\;3\; cm \quad (c)\;2\; cm \quad (d)\;1\;cm\]

Can you answer this question?

1 Answer

0 votes
Let 'u' be initial velocity
$\bigg(\large\frac{u}{2}\bigg)^2$$-u^2=-2a \times 3$
Since it loses half its velocity after 3 cm
or $a=\large\frac{u^2}{8}$
Let x be the further distance to which it penetrates
Therefore Total distance $=x'=3+x$
$0-u^2=-2 \large\frac{u^2}{8}$$x'$
Therefore $x=4-3=1\; cm$
Hence d is the correct answer.
answered Jun 28, 2013 by meena.p
edited Mar 26, 2014 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App