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
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K.E. of colliding body before collision $= \frac{1}{2} m v^{2}$
After collision its velocity becomes
$v' = \frac{(m_1 - m_2)}{(m_1 + m_2)} v = \frac{m}{3m} v = \frac{v}{3}$
∴ ∴ K.E. after collision $\frac{1}{2}mv'^{2}$ $= \frac{1}{2} mv^{2}$
Ratio of kinetic energy = $\frac{\text{K.E}_{\text{before}}}{\text{K.E}_{\text{after}}} = \frac{\frac{1}{2} mv^{2}}{\frac{1}{2} m \frac{v^{2}}{9}} = 9 : 1$
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