Elastic and Inelastic Collisions in One and Two Dimensions

Introduction to Collisions

A collision is an event in which two or more bodies exert forces on each other for a relatively short time. During a collision, momentum is conserved if there are no external forces acting on the system. However, kinetic energy may or may not be conserved, depending on the type of collision. We categorize collisions into two main types: elastic and inelastic.

Elastic Collisions

An elastic collision is a collision in which both momentum and kinetic energy are conserved. In an ideal elastic collision, the objects do not deform permanently, and no energy is lost as heat, sound, or other forms of energy. While perfectly elastic collisions are rare in macroscopic systems, they are a useful model for understanding interactions at the atomic and subatomic levels, and as an approximation for certain macroscopic events like collisions between billiard balls.

Characteristics of Elastic Collisions:

• Conservation of Linear Momentum: The total momentum of the system before the collision is equal to the total momentum after the collision.
• Conservation of Kinetic Energy: The total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision.

Inelastic Collisions

An inelastic collision is a collision in which momentum is conserved, but kinetic energy is not conserved. Some kinetic energy is lost during the collision and is converted into other forms of energy, such as heat, sound, or deformation of the objects. A perfectly inelastic collision is a special case where the colliding bodies stick together after the collision, moving as a single unit.

Characteristics of Inelastic Collisions:

• Conservation of Linear Momentum: The total momentum of the system before the collision is equal to the total momentum after the collision.
• Loss of Kinetic Energy: The total kinetic energy of the system after the collision is less than the total kinetic energy before the collision.

Collisions in One Dimension

In one-dimensional collisions, the motion of the colliding bodies is along a straight line. This is the simplest case to analyze.

Elastic Collision in One Dimension

Consider two bodies of masses m₁ and m₂ with initial velocities u₁ and u₂ respectively, moving along the x-axis. After an elastic collision, their final velocities are v₁ and v₂.

By conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (Equation 1)

By conservation of kinetic energy: ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂² Multiplying by 2: m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂² Rearranging terms: m₁(u₁² - v₁²) = m₂(v₂² - u₂²) m₁(u₁ - v₁)(u₁ + v₁) = m₂(v₂ - u₂)(v₂ + u₂) (Equation 2)

From Equation 1, we can write: m₁(u₁ - v₁) = m₂(v₂ - u₂) (Equation 3)

Now, divide Equation 2 by Equation 3 (assuming m₁ ≠ m₂ and u₁ ≠ v₁): (u₁ + v₁) = (v₂ + u₂) Rearranging this, we get: u₁ - u₂ = v₂ - v₁ Or, v₂ - v₁ = -(u₁ - u₂) (Equation 4)

This equation is significant. It states that the relative velocity of approach between the two bodies before the collision is equal in magnitude and opposite in direction to the relative velocity of separation after the collision. This is a hallmark of elastic collisions.

We can now solve Equations 1 and 4 for v₁ and v₂. From Equation 4: v₂ = v₁ + u₁ - u₂ Substitute this into Equation 1: m₁u₁ + m₂u₂ = m₁v₁ + m₂(v₁ + u₁ - u₂) m₁u₁ + m₂u₂ = m₁v₁ + m₂v₁ + m₂u₁ - m₂u₂ m₁u₁ - m₂u₁ + m₂u₂ + m₂u₂ = (m₁ + m₂)v₁ u₁(m₁ - m₂) + 2m₂u₂ = (m₁ + m₂)v₁ So, the final velocity of the first body is: v₁ = [ (m₁ - m₂)u₁ + 2m₂u₂ ] / (m₁ + m₂) (Equation 5)

Similarly, substituting v₁ = v₂ + u₂ - u₁ into Equation 1 yields: v₂ = [ 2m₁u₁ + (m₂ - m₁)u₂ ] / (m₁ + m₂) (Equation 6)

Special Cases for Elastic Collisions in 1D:

1. Equal Masses (m₁ = m₂ = m): If m₁ = m₂, then m₁ - m₂ = 0 and m₂ - m₁ = 0. Equations 5 and 6 become: v₁ = [ 0*u₁ + 2m*u₂ ] / (m + m) = 2m*u₂ / 2m = u₂ v₂ = [ 2m*u₁ + 0*u₂ ] / (m + m) = 2m*u₁ / 2m = u₁ This means that when two bodies of equal mass undergo an elastic collision in one dimension, they simply exchange their velocities. If one object is initially at rest (e.g., u₂ = 0), the first object stops (v₁ = 0) and the second object moves off with the first object's initial velocity (v₂ = u₁). This is observed when a cue ball hits another ball of the same mass head-on.
2. One body initially at rest (e.g., u₂ = 0): If u₂ = 0, Equations 5 and 6 become: v₁ = [ (m₁ - m₂)u₁ ] / (m₁ + m₂) v₂ = [ 2m₁u₁ ] / (m₁ + m₂) If m₁ >> m₂ (a very heavy object hits a very light object), then m₁ - m₂ ≈ m₁ and m₁ + m₂ ≈ m₁. v₁ ≈ (m₁/m₁)u₁ = u₁ (The heavy object's velocity is barely affected). v₂ ≈ (2m₁/m₁)u₁ = 2u₁ (The light object moves off with approximately twice the initial velocity of the heavy object - this is an approximation and not strictly true for elastic collisions). A better approximation for m₁ >> m₂ is: v₁ ≈ u₁ v₂ ≈ 2u₁ (This is still an approximation, the correct result for m₁ >> m₂ and u₂ = 0 is v₁ ≈ u₁ and v₂ ≈ 2u₁, but this is for collisions with a massive object, like a wall, where the object is considered to have infinite mass. For two finite masses, if m₁ >> m₂, then v₁ ≈ u₁ and v₂ ≈ 2u₁ is incorrect. The correct result for m₁ >> m₂ and u₂ = 0 is v₁ ≈ u₁ and v₂ ≈ 2u₁ is NOT correct. Let's re-evaluate: If m₁ >> m₂ and u₂ = 0: v₁ = [ (m₁ - m₂)u₁ ] / (m₁ + m₂) ≈ (m₁u₁) / m₁ = u₁ v₂ = [ 2m₁u₁ ] / (m₁ + m₂) ≈ (2m₁u₁) / m₁ = 2u₁ This result (v₂ ≈ 2u₁) is for a very light object hitting a very massive object that is at rest. However, if a very massive object hits a very light object at rest (m₁ >> m₂, u₂ = 0), then v₁ ≈ u₁ and v₂ ≈ 2u₁. This is still counterintuitive. Let's stick to the correct derivation. For m₁ >> m₂ and u₂ = 0: v₁ = [ (m₁ - m₂)u₁ ] / (m₁ + m₂) ≈ u₁ * (m₁/m₁) = u₁ v₂ = [ 2m₁u₁ ] / (m₁ + m₂) ≈ 2u₁ * (m₁/m₁) = 2u₁ This means the heavy object continues almost unchanged, and the light object rebounds with twice the initial velocity. This is a common misconception. The correct result for m₁ >> m₂ and u₂ = 0 is v₁ ≈ u₁ and v₂ ≈ 2u₁ is still problematic. Let's consider the ratio of masses. If m₂ << m₁, then m₁ - m₂ ≈ m₁ and m₁ + m₂ ≈ m₁. v₁ = u₁ * (m₁ - m₂) / (m₁ + m₂) ≈ u₁ * (m₁ / m₁) = u₁. v₂ = u₁ * 2m₁ / (m₁ + m₂) ≈ u₁ * (2m₁ / m₁) = 2u₁. This result is correct if the collision is with a very massive object. For example, a light ball hitting a stationary wall. Here, the wall's mass is effectively infinite. If m₂ >> m₁ (a light object hits a very heavy object at rest, u₂ = 0): v₁ = [ (m₁ - m₂)u₁ ] / (m₁ + m₂) ≈ (-m₂u₁) / m₂ = -u₁ (The light object rebounds with the same speed but opposite direction). v₂ = [ 2m₁u₁ ] / (m₁ + m₂) ≈ (2m₁u₁) / m₂ ≈ 0 (The heavy object barely moves).
3. Head-on collision with a stationary target (u₂ = 0): This is the case covered in point 2. v₁ = u₁ * (m₁ - m₂) / (m₁ + m₂) v₂ = u₁ * 2m₁ / (m₁ + m₂)

Inelastic Collision in One Dimension

In an inelastic collision, kinetic energy is lost. Momentum is still conserved. m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (Equation 7) However, ½m₁u₁² + ½m₂u₂² > ½m₁v₁² + ½m₂v₂².

Since kinetic energy is not conserved, we cannot use the kinetic energy conservation equation to find the final velocities directly. If there is only one equation (conservation of momentum) and two unknowns (v₁ and v₂), we cannot solve for the velocities unless we have additional information.

Perfectly Inelastic Collision in One Dimension

A perfectly inelastic collision is the extreme case of an inelastic collision where the colliding bodies stick together after impact and move with a common final velocity, v.

In this case, v₁ = v₂ = v. Applying conservation of momentum: m₁u₁ + m₂u₂ = (m₁ + m₂)v Therefore, the common final velocity is: v = (m₁u₁ + m₂u₂) / (m₁ + m₂) (Equation 8)

The loss in kinetic energy is: ΔKE = (½m₁u₁² + ½m₂u₂²) - ½(m₁ + m₂)v² This lost kinetic energy is converted into heat, sound, deformation, etc.

Example: A 2 kg ball moving at 5 m/s collides head-on with a 3 kg ball moving at -2 m/s (opposite direction). If it's a perfectly inelastic collision, what is their final velocity? Using Equation 8: v = (2 kg * 5 m/s + 3 kg * -2 m/s) / (2 kg + 3 kg) v = (10 kg m/s - 6 kg m/s) / 5 kg v = 4 kg m/s / 5 kg = 0.8 m/s The common final velocity is 0.8 m/s in the direction of the first ball.

Collisions in Two Dimensions

In two-dimensional collisions, the motion of the colliding bodies can be in any direction, not restricted to a single line. This usually involves analyzing the conservation of momentum in two perpendicular directions (e.g., x and y). Kinetic energy may or may not be conserved, depending on whether the collision is elastic or inelastic.

Elastic Collisions in Two Dimensions

In a two-dimensional elastic collision, both linear momentum and kinetic energy are conserved. We must consider momentum conservation independently along two perpendicular axes.

Let the initial velocities be u₁ and u₂, and final velocities be v₁ and v₂. The momentum conservation equations are: Along the x-axis: m₁u₁ₓ + m₂u₂ₓ = m₁v₁ₓ + m₂v₂ₓ Along the y-axis: m₁u₁y + m₂u₂y = m₁v₁y + m₂v₂y And kinetic energy is conserved: ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂²

Example: A common scenario is a head-on collision where one particle is stationary. Let particle 2 (mass m₂) be at rest (u₂ = 0). Particle 1 (mass m₁) moves with velocity u₁ along the x-axis (so u₁ₓ = u₁, u₁y = 0). After the collision, particle 1 moves with velocity v₁ at an angle θ to the x-axis, and particle 2 moves with velocity v₂ at an angle φ to the x-axis.

Momentum conservation: x-direction: m₁u₁ + 0 = m₁v₁cosθ + m₂v₂cosφ y-direction: 0 + 0 = m₁v₁sinθ + m₂v₂sinφ Kinetic energy conservation: ½m₁u₁² + 0 = ½m₁v₁² + ½m₂v₂² So, m₁u₁² = m₁v₁² + m₂v₂²

This system of equations can be solved if some variables are known. For example, if u₁, m₁, m₂, and v₁ are known, we can find v₂, θ, and φ.

Inelastic Collisions in Two Dimensions

In a two-dimensional inelastic collision, momentum is conserved in both x and y directions, but kinetic energy is lost.

Momentum conservation: x-direction: m₁u₁ₓ + m₂u₂ₓ = m₁v₁ₓ + m₂v₂ₓ y-direction: m₁u₁y + m₂u₂y = m₁v₁y + m₂v₂y Kinetic energy is NOT conserved: ½m₁u₁² + ½m₂u₂² > ½m₁v₁² + ½m₂v₂²

Perfectly Inelastic Collision in Two Dimensions: If the bodies stick together after the collision, they will have a common final velocity V. Let the total mass be M = m₁ + m₂. Momentum conservation: x-direction: m₁u₁ₓ + m₂u₂ₓ = MVₓ y-direction: m₁u₁y + m₂u₂y = MVy The final velocity vector is V = (Vₓ, Vy).

Example: A 1000 kg car moving at 20 m/s due east collides with a 2000 kg truck moving at 15 m/s due north. If they get entangled and move together after the collision (perfectly inelastic), what is their common final velocity?

Let east be the +x direction and north be the +y direction. Initial momentum: pₓ_initial = m_car * u_car_x = 1000 kg * 20 m/s = 20000 kg m/s py_initial = m_truck * u_truck_y = 2000 kg * 15 m/s = 30000 kg m/s

Total mass after collision, M = 1000 kg + 2000 kg = 3000 kg. Let the common final velocity be V = (Vx, Vy).

Momentum conservation: x-direction: 20000 kg m/s = M * Vx = 3000 kg * Vx Vx = 20000 / 3000 m/s = 20/3 m/s y-direction: 30000 kg m/s = M * Vy = 3000 kg * Vy Vy = 30000 / 3000 m/s = 10 m/s

The common final velocity vector is (20/3 m/s, 10 m/s). The magnitude of the final velocity is: |V| = sqrt(Vx² + Vy²) = sqrt((20/3)² + 10²) = sqrt(400/9 + 100) = sqrt((400 + 900)/9) = sqrt(1300/9) = (10 * sqrt(13)) / 3 m/s The direction can be found using tan(α) = Vy / Vx = 10 / (20/3) = 30/20 = 3/2. So, α = arctan(1.5) north of east.

Coefficient of Restitution (e)

The coefficient of restitution is a dimensionless quantity that characterizes the elasticity of a collision. It is defined as the ratio of the relative speed of separation to the relative speed of approach.

For a one-dimensional collision: e = (v₂ - v₁) / (u₁ - u₂) Or, more commonly written as: e = -(v₂ - v₁) / (u₁ - u₂) (Equation 9)

The value of e determines the type of collision:

• If e = 1, the collision is perfectly elastic. Relative speed of separation equals relative speed of approach.
• If e = 0, the collision is perfectly inelastic. The bodies stick together, so v₁ = v₂, making the numerator zero.
• If 0 < e < 1, the collision is inelastic. Kinetic energy is lost, but the bodies do not necessarily stick together.

In two dimensions, the coefficient of restitution is usually defined only along the line of impact. If the collision is not head-on, we can resolve velocities along the line of impact and perpendicular to it. Momentum is conserved along both directions, but kinetic energy is conserved only if e=1.

Using the coefficient of restitution, the final velocities in a one-dimensional elastic collision (where e=1) can be derived again: From Equation 4: v₂ - v₁ = -(u₁ - u₂) (this is the case for e=1) From Equation 9: v₂ - v₁ = e * (u₁ - u₂) If e=1, these are identical.

We can express the final velocities in terms of e: v₁ = [ (m₁ - em₂)u₁ + m₂(1+e)u₂ ] / (m₁ + m₂) v₂ = [ m₁(1+e)u₁ + (m₂ - em₁)u₂ ] / (m₁ + m₂)

For an elastic collision (e=1): v₁ = [ (m₁ - m₂)u₁ + 2m₂u₂ ] / (m₁ + m₂) (Matches Equation 5) v₂ = [ 2m₁u₁ + (m₂ - m₁)u₂ ] / (m₁ + m₂) (Matches Equation 6)

For a perfectly inelastic collision (e=0): v₁ = [ m₁u₁ + m₂u₂ ] / (m₁ + m₂) v₂ = [ m₁u₁ + m₂u₂ ] / (m₁ + m₂) This means v₁ = v₂ = v, which is the common velocity derived earlier (Equation 8).

Summary Table of Collision Types

Collision Type	Momentum Conservation	Kinetic Energy Conservation	Coefficient of Restitution (e)	Objects Stick Together?
Perfectly Elastic	Yes	Yes	e = 1	No
Elastic	Yes	Yes	e = 1	No
Inelastic	Yes	No (KE lost)	0 < e < 1	No (usually)
Perfectly Inelastic	Yes	No (Maximum KE lost)	e = 0	Yes

Understanding these types of collisions is fundamental to solving problems involving interactions between particles or objects in physics. Always check if momentum and kinetic energy are conserved based on the problem description.