Two masses $m_1$ and $m_2$ moving with velocities $V_1$ and $V_2$ in opposite directions collide elastically and after collision $m_1$ and $m_2$ move with velocities $V_2$ and $V_1$ respectively. The ratio $\dfrac{m_2}{m_1}$ is

By the principle of conservation of linear momentum:

$m_1 \vec{V}_1 + m_2 \vec{V}_2 = m_1 \vec{V}_1' + m_2 \vec{V}_2'$

Given that the final velocities are $\vec{V}_1' = \vec{V}_2$ and $\vec{V}_2' = \vec{V}_1$, we substitute these into the momentum equation:

$m_1 \vec{V}_1 + m_2 \vec{V}_2 = m_1 \vec{V}_2 + m_2 \vec{V}_1$

Rearranging the terms yields:

$m_1 (\vec{V}_1 - \vec{V}_2) = m_2 (\vec{V}_1 - \vec{V}_2)$

Since the masses are moving in opposite directions, $\vec{V}_1 \neq \vec{V}_2$, which implies $\vec{V}_1 - \vec{V}_2 \neq 0$.

Dividing both sides by $(\vec{V}_1 - \vec{V}_2)$ gives:

$m_1 = m_2$

Therefore, the ratio $\dfrac{m_2}{m_1} = 1.0$.

Note: In a perfectly elastic collision in one dimension, the velocities of the two bodies are exchanged if and only if their masses are equal.

Answer: $1.0$