Question

Two blocks of wood A and B are coupled by a spring and are placed on a frictionless table. If the spring is compressed and released, then $\frac{\text{kinetic energy of A}}{\text{kinetic energy of B}}=$

• A. $\frac{\text{mass of B}}{\text{mass of A}}$
• B. $\sqrt{\frac{\text{mass of B}}{\text{mass of A}}}$
• C. $\sqrt{\frac{\text{mass of A}}{\text{mass of B}}}$
• D. $\frac{\text{mass of A}}{\text{mass of B}}$

Answer 1

The correct option is A $\frac{\text{mass of B}}{\text{mass of A}}$

By law of conservation of momentum, we have

$m_A{v}_A+m_B{v}_B=0$

$m_A{v}_A=-m_B{v}_B$

$\frac{v_A}{v_B}=-\frac{m_B}{m_A}$

Now, findind the ratio of kinetic energies, we have

$\frac{K_A}{K_B}=\frac{\frac{1}{2}{m}_A{v}_A^2}{\frac{1}{2}{m}_B{v}_B^2}$

$=\frac{m_A}{m_B}\left(\frac{v_A^2}{v_B^2}\right)$

$=\left(\frac{m_A}{m_B}\right)(-\frac{m_B}{m_A}{)}^2$

$=\frac{m_B}{m_A}$