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 solve this problem, we need to understand the concept of kinetic energy and the conservation of momentum in a system where a stationary particle breaks into two parts.

1. The kinetic energy (\(K\)) of an object with mass \(m\) moving at velocity \(v\) is given by the formula: \(K = \frac{1}{2} m v^2\).
2. Let's denote the kinetic energies of the two parts as \(K_A\) and \(K_B\), for masses \(m_A\) and \(m_B\) moving at 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. We need to find the ratio of the kinetic energies, \(K_B : K_A\). Thus:

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

1. Applying the principle of conservation of momentum, where the total initial momentum is zero (since the particle is initially stationary), we have: \(m_A v_A = m_B v_B\).
2. Using the above momentum relation, we can express: \(v_A = \frac{m_B v_B}{m_A}\).
3. Substituting this into the kinetic energy ratio equation gives:

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

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

Answer 2

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.