Question

Obtain the expressions for the velocities of the two bodies after collision in the case of one dimensional elastic collision and discuss the special cases.

Answer 1

Consider two elastic bodies of masses m₁ and m₂ moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure.

In order to have collision, we assume that the mass m₁ moves faster than mass m₂ i.e., u₁ > u₂. 

For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same

From the law of conservation of linear momentum, Total momentum before collision (p_i) = Total momentum after collision (p_f)

For elastic collision, 

Total kinetic energy before collision KE_i = Total kinetic energy after collision KF_f

After simplifying and rearranging the terms,

Using the formula a² – b² = (a + b) (a – b), we can rewrite the above equation as

m₁ (u₁ + v₁ )(u₁ – v₁ ) = m₂ (v₂ + u₂ ) (v₂ – u₂ ) …(iv)

Dividing equation (iv) by (ii) gives,

This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.

Rewriting the above equation for V₁ and V_2,

v₁ = v₂ + u₂ – u₁ 

Or v₂ = u₁ + v₁ – u₁  ............. (vii)

To find the final velocities v₁ and v₂ : 

Substituting equation (vii) in equation (ii) gives the velocity of m₁ as

Similarly, by substituting (vi) in equation (ii) or substituting equation (viii) in equation (vii), 

we get the final velocity of m₂ as

Case 1: 

When bodies has the same mass i.e., m₁ = m₂ ,

The equations (x) and (xi) show that in one dimensional elastic collision, when two bodies of equal mass collide after the collision their velocities are exchanged.

Case 2:

When bodies have the same mass i.e., m₁ = m₂ and second body (usually called target) is at rest (u = 0),

By substituting m₁ m₂ = and u₂ = 0 in equations (viii) and equations (ix) we get, from equation (viii) ⇒ V₁ = 0 …(xii)

from equation (ix) ⇒ v₂ = u₁ … (xiii) Equations (xii) and (xiii) show that when the first body comes to rest the second body moves with the initial velocity of the first body.

Case 3:

The first body is very much lighter than the second body

Similarly, Dividing numerator and denominator of equation (ix) by m₂, we get

v₂ = 0

The equation (xiv) implies that the first body which is lighter returns back (rebounds) in the opposite direction with the same initial velocity as it has a negative sign. The equation (xv) implies that the second body which is heavier in mass continues to remain at rest even after collision. For example, if a ball is thrown at a fixed wall, the ball will bounce back from the wall with the same velocity with which it was thrown but in opposite direction.

Similarly, Dividing numerator and denominator of equation (xiii) by m₁, we get

The equation (xvi) implies that the first body which is heavier continues to move with the same initial velocity. The equation (xvii) suggests that the second body which is lighter will move with twice the initial velocity of the first body. It means that the lighter body is thrown away from the point of collision.