In a completely inelastic collision, the two bodies stick together and move with a common final velocity. Here, before the collision, body $A$ is moving with a velocity $v_1$ and body $B$ is at rest. The conservation of momentum must hold true because no external forces are acting on the system.

The momentum before the collision is only due to body $A$ since body $B$ is at rest. Therefore, the total initial momentum $p_{\text{initial}}$ of the system is given by:

$$p_{\text{initial}} = m_A v_1 + m_B v_0 = mv_1 + 0 = mv_1$$

where:

• $m_A$ and $m_B$ are the masses of bodies $A$ and $B$ respectively,
• $m$ is the mass of each body,
• $v_1$ is the velocity of body $A$,
• $v_0 = 0$ is the velocity of body $B$ because it is initially at rest.

Since they undergo a completely inelastic collision, bodies $A$ and $B$ stick together after the collision and hence move with a common velocity $v_2$. The total mass of the combined system post-collision is $$m_A + m_B = m + m = 2m$$. The momentum after the collision is given by:

$$p_{\text{final}} = (m_A + m_B)v_2 = 2m v_2$$

Applying the conservation of momentum (since no external force implies momentum is conserved), we equate $p_{\text{initial}}$ and $p_{\text{final}}$:

$mv_1 = 2m v_2$

Dividing through by $m$ yields:

$v_1 = 2v_2$

Thus, solving for the ratio $\frac{v_1}{v_2}$ gives:

$\frac{v_1}{v_2} = 2$

Therefore, the ratio of $v_1$ to $v_2$ is $2:1$, making the correct answer:

Option B: $2:1$