Question 63: 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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