Question 66: Centre of Mass and Rotational Motion

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 $\left(K_B: K_A\right)$ is :

Accepted Answer

A stationary particle breaks into two parts with masses $m_A$ and $m_B$, which then move with velocities $v_A$ and $v_B$, respectively. We need to determine the ratio of their kinetic energies $K_A$ and $K_B$.
Since the initial momentum of the particle is zero, the momentum of the two parts must be equal and opposite to conserve momentum:
$ m_A v_A = m_B v_B $
Here, the ratio of kinetic energies is given by:
$ \frac{K_A}{K_B} = \frac{\frac{1}{2} m_A v_A^2}{\frac{1}{2} m_B v_B^2} = \frac{m_A v_A^2}{m_B v_B^2} $
However, using the momentum relationship, we can substitute $m_A v_A = m_B v_B$ into the kinetic energy ratio, leading us to simplify:
$ \frac{K_A}{K_B} = \frac{v_A}{v_B} $
Therefore, the ratio of their kinetic energies $\left(K_B: K_A\right)$ is:
$ \frac{K_B}{K_A} = \frac{v_B}{v_A} $

Answer

$v_B: v_A$

Answer

$1: 1$

Answer

$m_B v_B: m_A v_A$

Answer

$m_B: m_A$

Centre of Mass and Rotational Motion

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 \left(K_B: K_A\right) is :

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