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

To resolve this matter, an understanding of kinetic energy and the conservation of momentum within a system where a static particle disintegrates into two components is requisite.

1. The kinetic energy (\(K\)) for an entity possessing mass \(m\) and traversing at velocity \(v\) is quantified by the expression: \(K = \frac{1}{2} m v^2\).
2. Let the kinetic energies of the two resulting fragments be designated as \(K_A\) and \(K_B\), corresponding to masses \(m_A\) and \(m_B\) and velocities \(v_A\) and \(v_B\) respectively: \(K_A = \frac{1}{2} m_A v_A^2\) and \(K_B = \frac{1}{2} m_B v_B^2\).
3. The objective is to determine the ratio \(K_B : K_A\). Consequently:

\(\frac{K_B}{K_A} = \frac{\frac{1}{2} m_B v_B^2}{\frac{1}{2} m_A v_A^2} = \frac{m_B v_B^2}{m_A v_A^2}\)

1. Adhering to the principle of momentum conservation, where the initial total momentum is null (given the particle's initial stationary state), we establish: \(m_A v_A = m_B v_B\).
2. Leveraging the aforementioned momentum relationship, we can express: \(v_A = \frac{m_B v_B}{m_A}\).
3. Substituting this derived velocity into the kinetic energy ratio equation yields:

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

1. Thus, the ratio of their kinetic energies is \(v_B : v_A\).