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:
- \(v_B : v_A\)
- \(m_B : m_A\)
- \(m_B v_B : m_A v_A\)
- \(1 : 1\)
The Correct Option is A
Solution and Explanation
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: vB : vA.
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