Question

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

Answer 1

Answer: (D)

K. E. of colliding body before collision $=\frac{1}{2} m v^2$
After collision its velocity becomes
$v^{\prime}=\frac{\left(m_1-m_2\right)}{\left(m_1+m_2\right)} v=\frac{m}{3 m} v=\frac{v}{3}$
$\therefore$ K. E. after collision $=\frac{1}{2} \cdot \frac{m v^2}{9}$
Ratio of kinetic energy $=\frac{K \cdot E_{\text {before }}}{K \cdot E_{\text {after }}}=\frac{\frac{1}{2} m v^2}{\frac{1 m v^2}{2 \times 9}}=9: 1$