Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity v₁ while body B is at rest before collision. The velocity of the system after collision is v₂. The ratio v₁ : v₂ is

The correct option is 2:1.

Let us solve the problem step-by-step using the principle of conservation of linear momentum.

Step 1: Identify the given information
- Mass of body A, $m_A=\; m$
- Mass of body B, $m_B=\; m$ (since they have the same mass)
- Initial velocity of body A before collision, $u_A=\; v$₁
- Initial velocity of body B before collision, $u_B=\; 0$ (since B is at rest)
- Velocity of the combined system after the completely inelastic collision, $v_{\text{final}}=\; v$₂

Step 2: Apply the Principle of Conservation of Linear Momentum
According to the law of conservation of momentum, the total momentum before collision must be equal to the total momentum after collision.
For a completely inelastic collision, the two bodies stick together and move with a common velocity after the collision.
Therefore, the equation is:
$m_A{u}_A+m_B{u}_B=\; (m$_A + m_B)v₂

Substituting the given values into the equation:
$m(v$₁) + m(0) = (m + m)v₂
$m\; v$₁ = 2m v₂

Step 3: Calculate the ratio v₁ : v₂
We can divide both sides of the equation by $m$ (since $m\; \ne \; 0$):
$v$₁ = 2v₂
Rearranging the terms to find the ratio $v$₁ : v₂:
$\frac{v_1}{v_2}=\; 2$
This can be written as:
$v_1:v_2=\; 2\; :\; 1$

Thus, the ratio of the initial velocity of body A ($v$₁) to the velocity of the combined system after collision ($v$₂) is 2:1.