Question

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 :

• A. 1 : 2
• B. 2 : 1
• C. 4 : 1
• D. 1 : 4

Answer 1

The Correct Option is B

To solve the problem of determining the ratio \(v_1 : v_2\) for a completely inelastic collision between two bodies A and B, both having the same mass, let's apply the principle of conservation of momentum. In such a collision, the two bodies stick together and move with a common velocity after the collision.

Given: Mass of A = \(m\), Velocity of A before collision = \(v_1\), Mass of B = \(m\), Velocity of B before collision = \(0\), Combined velocity after collision = \(v_2\).

Conservation of momentum: The total momentum before the collision equals the total momentum after the collision.

Before collision, total momentum = \(m \times v_1 + m \times 0 = mv_1\).

After collision, total momentum = \( (m + m) \times v_2 = 2mv_2\).

Equation:

\(mv_1 = 2mv_2\)

Canceling the mass \(m\) from both sides, we have:

\(v_1 = 2v_2\)

Ratio \(v_1:v_2\):

\(v_1 : v_2 = 2 : 1\)

Thus, the correct answer is the ratio 2 : 1.