Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
0 votes

A body of mass m moving with velocity v collides head on with another body of mass $2m$ which is initially at rest. The ratio of K.E of colliding body before and after collision will be

\[(a)\;1:1 \quad (b)\;2:1\quad (c)\; 4:1 \quad (d)\;9:1\]
Can you answer this question?

1 Answer

0 votes
$KE$ of colliding bodies before collision $=\large\frac{1}{2} $$mv^2$
After collision the mass $=m+2m=3m$
Velocity becomes $v'=\bigg(\large \frac{m_!-m_2}{m_1+m_2}\bigg)v$
$\qquad= \large\frac{m}{3m} $$v=\large\frac{v}{3}$
KE after collision $=\large\frac{1}{2} $$m \bigg(\large\frac{v}{3}\bigg)^2$
$\qquad=\large\frac{1}{9} \frac{1}{2}$$ mv^2$
$\large\frac{KE _{before}}{KE_{after}}=\large\frac{\Large\frac{1}{2} \large mv^2}{\Large\frac{1}{9} \times \frac{1}{2} \large mv^2}$
Hence d is the correct answer.


answered Aug 9, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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