Question

Derive an expression for the speeds of two colliding bodies having one dimensional elastic collision.

Answer 1

Let two bodies A and B of masses m₁ and m₂ are moving in a straight line with velocities u₁ and u₂ respectively (u₁ > u₂) After collision, let their velocities be changed to v₁ and v₂ respectively. (Fig above)

Total linear momentum of the system (bodies A & B) before collision = m₁u₁ + m₂u₂

Total linear momentum of the system (bodies A and B) after collision = m₁v₁ + m₂v₂

According to the law low of conservation of linear momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂   ...(i)

or, m₁(u₁ - u₂) = m₂(v₂ - u₂)    ....(ii)

As the collision is elastic, so the kinetic energy of the system also remains conserved:

i.e., 1/2 m₁u₁² + 1/2 m₂u₂² = 1/2 m₁v₁² + 1/2 m₂v₂²   ....(iii)

m₁(u₁² - v₁²) = m₂(v₂² - u₂²)

or, m₁(u₁² - v₁²)(u₁ - v₁) = m₂(v₂ - u₂)(v₂ - u₂)  ...(iv)

Dividing equation (iv) by equation (ii) we get

u₁ + v₁ = v₂ + u₂   ...(v)

or, u₁ + u₂ = v₂ + v₁   ...(vi)

or, Relative velocity of approach = Relative velocity of separation. That is, in an elastic one-dimensional collision, the relative velocity of approach before collision is equal to the relative velocity of separation after the collision.

The ratio of the relative velocity of separation to the relative velocity of approach is called coefficient of restitution (e), i.e.,

e = {v₂ - v₁}/{u₁ - u₂}   ...(vii)

The value of e depends upon the nature of materials of the colliding bodies.

For perfectly elastic collision, e = 1

Velocities after collision:

From equation (vi), v₂ = u₁ - u₂ + v₁   ...(viii)

Substituting the value of v₂ in equation (ii), we have,

m₁(u₁ - v₁) = m₂[u₁ - 2u₂ + v₁]

or, v₁ = {2m₂u₂ + u₁(m₁ - m₂)}/{(m₁m₂)}  ...(ix)

Substituting the value of v₁ in equation (viii), we have,

v₂ = {2m₁u₁ + u₂(m₂ - m₁)}/{m₁ + m₂}    ...(x)

Special cases:

(1) When the two bodies have equal masses i.e., m₁ = m₂ = m (say).

From equation (ix), we get

and from equation (ix), we get

It follows from equation (xi) and (xii) that if two bodies of equal masses undergo one dimensional elastic collision, then their velocities are swapped after collision.

(2) When u₂ = 0, i.e., body B is at rest

In such a situation, equation (ix) and (x) reduce to;

Three sub-cases arise:

(i) When m₁ = m₂, i.e., both bodies A and B have same mass.

Then from equations (xiii) and (xiv), we have v₁ = 0 and v₂ = u₁

Thus, in an elastic collision in one dimension, when a body collides with another body at rest and of same mass, it is stopped and the other body begins to move with the velocity of the first.

(ii) When m₁ > m₂, i.e., body B is lighter than the body A. In such a situation, m₂ can neglected as compared to m₁

Thus, if a massive body collides with a lighter body at rest, there is practically no change in its velocity, whereas the stationary body begins to move with twice the velocity of the first.

(iii) When m₁ << m₂, i.e., body A is lighter than body B. In such a situation, m₁ can be neglected as compared to m₂

Thus, if a light body collides elastically with a heavy body at rest, it rebounds with its original velocity, whereas the heavy body practically remains at rest.