logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Ad
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'$
$x'=4$
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
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...