Question

Arrive at an expression for elastic collision in one dimension and discuss various 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 the figure given below.

Elastic collision in one dimension

Mass	Initial velocity	Final velocity
Mass m1	u1	v1
Mass m2	u2	v2

In order to have a collision, we assume that the mass m] 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 a collision must remain the same.

	The momentum of mass m1	Momentum of mass m2	Total linear momentum
Before collision	Pi1 = m1u1	Pi2 = m2u2	`P_i = P_{i1} + P_{i2}`
			`P_i = m_1u_1 + m_2u_2`
After Collision	Pf1 = m1v1	Pf2 = m2v2	`P_f = P_{f1} + P_{f2}`
			`P_f = m_1v_1 + m_2v_2`

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

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ …(i)

Or `m_1(u_1 - v_1) = m_2(v_2 - u_2)` ..................(ii)

Further,

	Kinetic energy of mass `m_1`	Kinetic energy of mass `m_2`	Total kinetic energy
Before collision	`KE_{i1} = 1/2m_1u_1^2`	`KE_{i2} = 1/2m_2u_2^2`	`KE_i = KE_{i1} + KE_{i2}` `KE_i = 1/2m_1u_1^2 + 1/2m_2u_2^2`
After collision	`KE_{f1} = 1/2m_1v_1^2`	`KE_{f2} = 1/2m_2v_2^2`	`KE_i = KE_{i1} + KE_{i2}` `KE_f = 1/2m_1v_1^2 + 1/2m_2v_2^2`

For elastic collision,

Total kinetic energy before collision `KE_i` = Total kinetic energy after collision `KE_f`

`1/2m_1u_1^2 + 1/2m_2u_2^2 = 1/2m_1v_1^2 + 1/2m_2v_2^2` .....................(iii)

After simplifying and rearranging the terms,

`m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2)`

Using the formula `a^2 - b^2` = (a + b)(a - b), we can rewrite the above equation as

`m_1(u_1 + v_1)(u_1 - v_1) = m_2(v_2 + u_2)(v_2 - u_2)` .......................(iv)

Diving equation (iv) by (ii) gives,

`(m_1(u_1 + v_1)(u_1 - v_1))/(m_1(u_1 - v_1)) = (m_2(v_2 + u_2)(v_2 - u_2))/(m_2(v_2 - u_2))`

`u_1 + v_1 = v_2 + u_2`

`u_1 - u_2 = v_2 - v_1` .............Rearranging, (v)

Equation (v) can be rewritten as

`u_1 - u_2 = -(v_1 - v_2)`

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 v₂,
v₁ = v₂ + u₂ – u₂ …(vi)

Or v₂ = u₁ + v₁ – u₂ …(vii)

To find the final velocities v₁ and v₂:

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

m₁ (u₁ – v₁) = m₂(u₁ + v₁ – u₂ – u₂)

m₁ (u₁ – y₁) = m₂ (u₁ + + v₁ – 2u₂)

m₁u₁ – m₁v₁ = m₂u₁ + m₂v₁ + 2m₂u₂

m₁u₁ – m₂u₁ + 2m₂u₂ = m₁v₁ + m₂v₁

(m₁– m₂) u₁ + 2m₂u₂ = (m₁ + m₂) v₁

Or `v_1 = ((m_1 - m_2)/(m_1 + m_2))u_1 + ((2m_2)/(m_1 + m_2))u_2` ................(viii)

Similarly, by substituting (vi) in equation (ii) or substituting equation (viii) in equation (vii), we get the final velocity of m₂ as

`v_2 = ((2m_1)/(m_1 + m_2))u_1 + ((m_2 - m_1)/(m_1 + m_2))u_2` ....................(ix)

Case 1: When bodies has the same mass i.e., m₁ = m₂,

equation (viii) ⇒ `v_1 = (0)u_1 + ((2m_2)/(2m_2))u_2`

`v_1 = u_2` .............................(x)

equation (ix) ⇒ `v_2 = ((2m_1)/(2m_1))u_1 + (0)u_2`

`v_2 = u_1`  .............................(xi)

The equations (x) and (xi) show that in a 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 (u₂ = 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

(m₁ << m₂, `m_1/m_2` << 1) then the ratio `m_1/m_2 ≈ 0` and also if the target is at rest (`u_2 = 0`)

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

`v_1 = ((m_1/m_2 - 1)/(m_1/m_2 + 1))u_1 + (2/(m_1/m_2 + 1))`(0)

`v_1 = ((0 - 1)/(0 + 1))u_1`

`v_1 = -u_1`

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

`v_2 = ((2m_1/m_2)/(m_1/m_2 + 1))u_1 + ((1 - m_1/m_2)/(m_1/m_2 + 1))`(0)

`v_2 = (0)u_1 + ((1 - m_1/m_2)/(m_1/m_2 + 1))`(0)

`v_2 = 0` ................................(xv)

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 the 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.

Case 4: The second body is very much lighter than the first body (m₂ << m₁, `m_2/m_1` << 1) then the ratio `m_2/m_1 = 0` and also if the target is at rest `u_2 = 0`

Dividing numerator and denominator of equation (viii) by `m_1`, we get

`v_1 = ((1 - m_2/m_1)/(1 + m_2/m_1))u_1 + ((2m_2/m_1)/(1 + m_2/m_1))`(0)

`v_1 = ((1 - 0)/(1 + 0))u_1 + (0/(1 + 0))`(0)

`v_1 = u_1` .............(xvi)

Similarly,

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

`v_2 = (2/(1 + m_2/m_1))u_1 + ((m_2/m_1 - 1)/(1 + m_2/m_1))`(0)

`v_2 = (2/(1 + 0))u_1`

`v_2 = 2u_1` ...........(xvii)

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.