Question

A body of mass m moving with velocity v makes a head-on collision with another body of mass 2m which is initially at rest. The ratio of kinetic energies of colliding body before and after collision will be

• A. 9:1
• B. 1:1
• C. 4:!
• D. 2:!

Answer 1

The correct option is A 9:1

$\frac{K{E}_{before}}{k{E}_{after}}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{v}_1^2}=\frac{V^2}{V_1^2}$
Since $V_1=\frac{m_1-m_2}{m_1+m_2}{V}_1=\frac{m-2m}{m+2m}V=-\frac{V}{3}$
Hence ratio of KE$=\frac{V^2}{V_1^2}=9:1$