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

Answer: (A)

$v_B : v_A$

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 : vA.