Question

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

• A. 1 : 1
• B. 2 : 1
• C. 4 : 1
• D. 9 : 1

Answer 1

The correct option is D

9 : 1

K.E. of colliding body before collision $=\frac{1}{2}m{v}^2$

After collision its velocity becomes $v^{\prime }=\frac{(m_1-m_2)}{(m_1+m_2)}v=\frac{m}{3m}v=\frac{v}{3}$

$\therefore$ K.E. after collision $\frac{1}{2}m{v}^{\prime 2}=\frac{1}{2}\frac{m{v}^2}{9}$

Ratio of kinetic energy $=\frac{K.E{.}_{before}}{K.E{.}_{after}}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}\frac{m{v}^2}{9}}=9:1$