Define elastic collision. Show that in one dimensional elastic collision of two bodies, the relative velocity of separation after collision is equal to the relative velocity of approach before collision.

In physics, an elastic collision refers to a type of collision between two objects in which both the momentum and kinetic energy are conserved. In such a collision, the total momentum of the system before and after the collision remains the same, as does the total kinetic energy.

Now, let's consider a one-dimensional elastic collision between two bodies, labeled as body 1 and body 2. Before the collision, the two bodies are approaching each other with certain velocities, and after the collision, they separate with new velocities.

Let the initial velocities of body 1 and body 2 be denoted as u₁ and u₂, respectively. The final velocities after the collision are v₁ and v₂ for body 1 and body 2, respectively.

The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. In one dimension, this can be expressed as:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where m₁ and m₂ are the masses of body 1 and body 2, respectively.

Additionally, in an elastic collision, the total kinetic energy before and after the collision is conserved. The kinetic energy of an object is given by (1/2)mv², where m is the mass and v is the velocity. Thus, the conservation of kinetic energy can be expressed as:

(1/2)m₁u₁² + (1/2)m₂u₂² = (1/2)m₁v₁² + (1/2)m₂v₂²

To prove that the relative velocity of separation after the collision is equal to the relative velocity of approach before the collision, we need to show that v₂ - v₁ = u₁ - u₂.

Rearranging the momentum conservation equation, we have:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
=> m₁u₁ - m₁v₁ = m₂v₂ - m₂u₂
=> m₁(u₁ - v₁) = m₂(v₂ - u₂)

Now, rearranging the kinetic energy conservation equation, we have:

(1/2)m₁u₁² + (1/2)m₂u₂² = (1/2)m₁v₁² + (1/2)m₂v₂²
=> m₁u₁² - m₁v₁² = m₂v₂² - m₂u₂²
=> m₁(u₁ - v₁)(u₁ + v₁) = m₂(v₂ - u₂)(v₂ + u₂)

Dividing the momentum conservation equation by the kinetic energy conservation equation, we get:

(u₁ + v₁)/(u₁ - v₁) = (v₂ + u₂)/(v₂ - u₂)

Now, we can solve for v₂ - v₁:

(u₁ + v₁)/(u₁ - v₁) = (v₂ + u₂)/(v₂ - u₂)
(u₁ + v₁)(v₂ - u₂) = (v₂ + u₂)(u₁ - v₁)
u₁v₂ + v₁v₂ - u₂v₂ - u₁u₂ = u₁v₂ - v₁u₂ + u₂v₂ - u₁u₂
v₁v₂ - u₂v₂ = - v₁u₂ + u₂v₂
v₁v₂ - v₂u₂ = - v₁u₂ + u₂v₂
v₁(v₂ - u₂) = - u₂(v₂ - u₂)
v₁ = -u₂

Hence, we have shown that v₁