Question

A stationary particle breaks into two parts of masses \(m_A\) and \(m_B\), which move with velocities \(v_A\) and \(v_B\), respectively. The ratio of their kinetic energies (\(K_B : K_A\)) is:

• A. \(v_B : v_A\)
• B. \(m_B : m_A\)
• C. \(m_B v_B : m_A v_A\)
• D. \(1 : 1\)

Answer 1

The Correct Option is A

Initial momentum is zero:

\[ P_A = P_B \implies m_A v_A = m_B v_B. \]

The kinetic energy ratio is:

\[ \frac{K_B}{K_A} = \frac{\frac{1}{2} m_B v_B^2}{\frac{1}{2} m_A v_A^2}. \]
\[ \frac{K_B}{K_A} = \frac{m_B v_B^2}{m_A v_A^2} = \frac{v_B}{v_A}. \]

Final Answer: v_B : v_A.