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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

\[(a)\;1:1 \quad (b)\;2:1\quad (c)\; 4:1 \quad (d)\;9:1\]

1 Answer

$KE$ 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

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