Physics HRK Collisions: Complete Guide to Elastic & Inelastic Collisions with Impulse-Momentum Theorem

House of Physics October 19, 2025

Complete Guide to Collisions in Physics

Master Elastic and Inelastic Collisions, Impulse-Momentum Theorem, Conservation Laws, and Problem Solving
Collision Physics Elastic Collision Inelastic Collision Impulse-Momentum Theorem Conservation of Momentum Reading Time: 25 min

📋 Table of Contents

📜 Historical Background

The study of collisions has been fundamental to physics since the Scientific Revolution:

  • René Descartes (1644): Proposed conservation of "quantity of motion" (momentum)
  • Christian Huygens (1669): First correct analysis of elastic collisions
  • Isaac Newton (1687): Formulated laws of motion and impact in Principia Mathematica
  • John Wallis (1671): First to distinguish between elastic and inelastic collisions

These developments laid the foundation for our modern understanding of collision physics and conservation laws.

Introduction to Collisions

🔬 What is a Collision?

A collision is an interaction between bodies that occurs in a time interval \( \Delta t \) that is negligible compared to the time during which we observe the system. We can also characterize a collision as an event where external forces acting on the system are negligible compared to impulsive collision forces.

Important Note: Collision does not necessarily mean physical contact between two bodies. For example, the scattering of alpha particles by atomic nuclei is considered a collision without physical contact.

📝 Key Characteristics of Collisions

  • Occur over very short time intervals
  • Involve large impulsive forces
  • External forces are negligible compared to collision forces
  • Momentum is always conserved (if no external forces)
  • Kinetic energy may or may not be conserved

Types of Collisions

Elastic Collision

A collision in which both momentum and kinetic energy are conserved. No energy is transformed into other forms such as heat, light, or sound.

  • Momentum conserved: ✓
  • Kinetic energy conserved: ✓
  • Objects separate after collision

Inelastic Collision

A collision in which momentum is conserved but kinetic energy is not conserved. Some kinetic energy is converted into other forms.

  • Momentum conserved: ✓
  • Kinetic energy conserved: ✗
  • Objects may stick together

Completely Inelastic Collision

A special case of inelastic collision where the colliding objects stick together and move as one object after collision.

  • Maximum kinetic energy loss
  • Objects move with common velocity
  • Momentum is conserved

Elastic Collisions

🔬 Elastic Collision Definition

In an elastic collision:

  • Total momentum before collision = Total momentum after collision
  • Total kinetic energy before collision = Total kinetic energy after collision
  • No permanent deformation occurs
  • No energy is lost as heat, sound, or light

Examples: Collisions between atomic particles, billiard balls (approximately), Newton's cradle.

Inelastic Collisions

🔬 Inelastic Collision Definition

In an inelastic collision:

  • Total momentum before collision = Total momentum after collision
  • Total kinetic energy before collision ≠ Total kinetic energy after collision
  • Some kinetic energy is converted to other forms
  • Permanent deformation may occur

Examples: Car crashes, clay balls sticking together, most real-world collisions.

Impulse and Momentum

🔬 Impulsive Forces

An impulsive force is a force that acts for a very short time interval during a collision. For example, when a ball is hit with a bat, an impulsive force is applied.

These forces are typically much larger than other forces acting on the system and cause significant changes in momentum in very short time intervals.

Impulse-Momentum Theorem

🧮 Derivation of Impulse-Momentum Theorem

Step 1: Newton's Second Law

\[ F = \frac{dP}{dt} \]

According to Newton's second law, force equals the rate of change of momentum.

Step 2: Change in Momentum

\[ d\overline{P} = \overline{F} dt \]

The change in momentum \( dP \) of a particle in time \( dt \) during which the force acts.

Step 3: Integration Over Time

\[ \int_{t_i}^{t_f} d\overline{P} = \int_{t_i}^{t_f} \overline{F} dt \]
\[ \overline{P}_f - \overline{P}_i = \int_{t_i}^{t_f} \overline{F} dt \]

Integrating between initial time \( t_i \) and final time \( t_f \).

Step 4: Definition of Impulse

\[ \overline{J} = \int_{t_i}^{t_f} \overline{F} dt \]

Impulse \( \overline{J} \) is defined as the product of force and time.

Step 5: Impulse-Momentum Theorem

\[ \overline{J} = \overline{P}_f - \overline{P}_i \]

The impulse of the net force acting on a particle during a given time interval equals the change in momentum of the particle during that interval.

📝 Key Points About Impulse

  • Impulse is a vector quantity with the same units as momentum (kg·m/s)
  • The magnitude of impulse is represented by the area under the F(t) curve
  • For a constant average force \( \overline{F} \), impulse is \( \overline{J} = \overline{F} \Delta t \)
  • Impulse-momentum theorem is a vector equation dealing with change of momentum
  • Work-energy theorem is a scalar equation dealing with change in kinetic energy

Conservation of Momentum During Collision

🧮 Conservation of Momentum Derivation

Step 1: System of Two Particles

Consider a system of two particles that collide with each other. During the collision, each particle exerts an impulsive force on the other.

\[ \overline{F}_{12} = -\overline{F}_{21} \]

According to Newton's third law, the forces are equal and opposite.

Step 2: Change in Momentum

\[ \Delta \overline{P}_1 = \int \overline{F}_{12} dt \]
\[ \Delta \overline{P}_2 = \int \overline{F}_{21} dt \]

Change in momentum of each particle due to collision forces.

Step 3: Total Change in Momentum

\[ \Delta \overline{P} = \Delta \overline{P}_1 + \Delta \overline{P}_2 \]
\[ = \int \overline{F}_{12} dt + \int \overline{F}_{21} dt \]
\[ = \int (\overline{F}_{12} + \overline{F}_{21}) dt \]
\[ = \int 0 dt \]
\[ = 0 \]

The total change in momentum of the system is zero.

Step 4: Conservation of Momentum

\[ \overline{P}_i = \overline{P}_f \]

The total momentum of the system before collision equals the total momentum after collision.

💡 Important Note

Momentum conservation during collisions holds even when external forces act on the system, provided these external forces are negligible compared to the impulsive collision forces. This is typically true for short-duration collisions.

Elastic Collision in One Dimension

📐 One-Dimensional Elastic Collision

[Diagram: Two masses m₁ and m₂ moving with velocities u₁ and u₂ before collision, and v₁ and v₂ after collision]

Consider two particles of masses m₁ and m₂ moving along a straight line with velocities u₁ and u₂ before collision, and v₁ and v₂ after collision.

Derivation of Final Velocities

🧮 Derivation of Final Velocities in 1D Elastic Collision

Step 1: Conservation of Momentum

\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
\[ m_1 (u_1 - v_1) = m_2 (v_2 - u_2) \quad \text{(1)} \]

Step 2: Conservation of Kinetic Energy

\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
\[ m_1 (u_1^2 - v_1^2) = m_2 (v_2^2 - u_2^2) \quad \text{(2)} \]

Step 3: Factor Equation (2)

\[ m_1 (u_1 - v_1)(u_1 + v_1) = m_2 (v_2 - u_2)(v_2 + u_2) \quad \text{(3)} \]

Step 4: Divide Equation (3) by Equation (1)

\[ u_1 + v_1 = v_2 + u_2 \]
\[ u_1 - u_2 = v_2 - v_1 \quad \text{(4)} \]

This shows that in an elastic collision, the relative velocity of approach equals the relative velocity of separation.

Step 5: Solve for Final Velocities

From equation (4):

\[ v_2 = v_1 + u_1 - u_2 \quad \text{(5)} \]

Substitute into momentum equation (1):

\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 (v_1 + u_1 - u_2) \]
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_1 + m_2 u_1 - m_2 u_2 \]
\[ m_1 u_1 + m_2 u_2 - m_2 u_1 + m_2 u_2 = (m_1 + m_2) v_1 \]
\[ (m_1 - m_2) u_1 + 2 m_2 u_2 = (m_1 + m_2) v_1 \]
\[ v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 + \frac{2 m_2}{m_1 + m_2} u_2 \quad \text{(6)} \]

Step 6: Solve for v₂

\[ v_2 = \frac{2 m_1}{m_1 + m_2} u_1 + \frac{m_2 - m_1}{m_1 + m_2} u_2 \quad \text{(7)} \]

📝 Final Velocities in 1D Elastic Collision

\[ v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 + \frac{2 m_2}{m_1 + m_2} u_2 \]
\[ v_2 = \frac{2 m_1}{m_1 + m_2} u_1 + \frac{m_2 - m_1}{m_1 + m_2} u_2 \]

Special Cases of Elastic Collisions

Case 1: Equal Masses (m₁ = m₂)

When the colliding particles have equal masses:

\[ v_1 = u_2 \]
\[ v_2 = u_1 \]

The particles exchange their velocities after collision.

Case 2: Target at Rest (u₂ = 0)

When the second particle is initially at rest:

\[ v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 \]
\[ v_2 = \frac{2 m_1}{m_1 + m_2} u_1 \]

Case 3: Light Particle on Heavy Target

When m₁ ≪ m₂ (light particle striking heavy stationary target):

\[ v_1 \approx -u_1 \]
\[ v_2 \approx 0 \]

The light particle rebounds with almost the same speed.

Case 4: Heavy Particle on Light Target

When m₁ ≫ m₂ (heavy particle striking light stationary target):

\[ v_1 \approx u_1 \]
\[ v_2 \approx 2u_1 \]

The heavy particle continues almost unaffected, while the light target moves with twice the speed.

Inelastic Collision in One Dimension

🔬 Completely Inelastic Collision

In a completely inelastic collision, the colliding objects stick together and move as a single object after collision. This results in maximum kinetic energy loss.

For two particles of masses m₁ and m₂ with initial velocities u₁ and u₂:

\[ m_1 u_1 + m_2 u_2 = (m_1 + m_2) V \]
\[ V = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \]

where V is the common velocity after collision.

🧮 Kinetic Energy Loss in Inelastic Collision

Step 1: Initial Kinetic Energy

\[ K_i = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \]

Step 2: Final Kinetic Energy

\[ K_f = \frac{1}{2} (m_1 + m_2) V^2 \]
\[ = \frac{1}{2} (m_1 + m_2) \left( \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \right)^2 \]
\[ = \frac{1}{2} \frac{(m_1 u_1 + m_2 u_2)^2}{m_1 + m_2} \]

Step 3: Energy Loss

\[ \Delta K = K_i - K_f \]
\[ = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 - \frac{1}{2} \frac{(m_1 u_1 + m_2 u_2)^2}{m_1 + m_2} \]
\[ = \frac{1}{2} \left[ \frac{m_1 u_1^2 (m_1 + m_2) + m_2 u_2^2 (m_1 + m_2) - (m_1 u_1 + m_2 u_2)^2}{m_1 + m_2} \right] \]
\[ = \frac{1}{2} \left[ \frac{m_1^2 u_1^2 + m_1 m_2 u_1^2 + m_1 m_2 u_2^2 + m_2^2 u_2^2 - m_1^2 u_1^2 - 2 m_1 m_2 u_1 u_2 - m_2^2 u_2^2}{m_1 + m_2} \right] \]
\[ = \frac{1}{2} \left[ \frac{m_1 m_2 (u_1^2 + u_2^2 - 2 u_1 u_2)}{m_1 + m_2} \right] \]
\[ = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (u_1 - u_2)^2 \]

📝 Energy Loss in Completely Inelastic Collision

\[ \Delta K = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (u_1 - u_2)^2 \]

💡 Important Insight

The energy loss in a completely inelastic collision depends on:

  • The product of the masses (m₁m₂)
  • The square of their relative velocity (u₁ - u₂)²
  • It is maximum when the masses are equal

This explains why car crashes are so destructive - the kinetic energy is converted into deformation, heat, and sound.

Two-Dimensional Collisions

🔬 Two-Dimensional Collisions

In two-dimensional collisions, the motion occurs in a plane. We need to apply conservation of momentum in both x and y directions. For elastic collisions, we also apply conservation of kinetic energy.

For two particles colliding in 2D:

  • Momentum conservation: \( m_1 \overline{u_1} + m_2 \overline{u_2} = m_1 \overline{v_1} + m_2 \overline{v_2} \)
  • Kinetic energy conservation (elastic): \( \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \)

Elastic Collisions in 2D

📐 Two-Dimensional Elastic Collision

[Diagram: Two masses colliding in 2D with angles θ and φ showing directions after collision]

For a stationary target (u₂ = 0), the final velocities make angles θ and φ with the initial direction.

🧮 2D Elastic Collision with Stationary Target

Step 1: Conservation of Momentum (x-direction)

\[ m_1 u_1 = m_1 v_1 \cos\theta + m_2 v_2 \cos\phi \quad \text{(1)} \]

Step 2: Conservation of Momentum (y-direction)

\[ 0 = m_1 v_1 \sin\theta - m_2 v_2 \sin\phi \quad \text{(2)} \]

Step 3: Conservation of Kinetic Energy

\[ \frac{1}{2} m_1 u_1^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \quad \text{(3)} \]

Step 4: Special Case - Equal Masses

When m₁ = m₂, the equations simplify:

\[ u_1 = v_1 \cos\theta + v_2 \cos\phi \quad \text{(1a)} \]
\[ 0 = v_1 \sin\theta - v_2 \sin\phi \quad \text{(2a)} \]
\[ u_1^2 = v_1^2 + v_2^2 \quad \text{(3a)} \]

Step 5: Relationship Between Angles

For equal masses, we can show that:

\[ \theta + \phi = 90^\circ \]

The two particles move at right angles to each other after collision.

Inelastic Collisions in 2D

🔬 2D Inelastic Collision

In two-dimensional inelastic collisions, only momentum is conserved. The analysis is similar to elastic collisions but without the energy conservation equation.

For a completely inelastic collision in 2D:

\[ m_1 \overline{u_1} + m_2 \overline{u_2} = (m_1 + m_2) \overline{V} \]

where \( \overline{V} \) is the common velocity vector after collision.

Solved Problems

Problem 1: Elastic Collision of Equal Masses

A ball of mass m moving with velocity u collides elastically with another ball of equal mass at rest. What are the final velocities of the balls?

Given: m₁ = m₂ = m, u₁ = u, u₂ = 0
Using the formulas for elastic collision:
\[ v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 + \frac{2 m_2}{m_1 + m_2} u_2 \]
\[ = \frac{m - m}{m + m} u + \frac{2 m}{m + m} \times 0 \]
\[ = 0 \]
\[ v_2 = \frac{2 m_1}{m_1 + m_2} u_1 + \frac{m_2 - m_1}{m_1 + m_2} u_2 \]
\[ = \frac{2 m}{m + m} u + \frac{m - m}{m + m} \times 0 \]
\[ = u \]
Answer: The first ball comes to rest, and the second ball moves with velocity u.
Problem 2: Completely Inelastic Collision

A car of mass 1000 kg moving at 20 m/s collides with a stationary car of mass 1500 kg. If they stick together after collision, find their common velocity and the energy lost in the collision.

Given: m₁ = 1000 kg, u₁ = 20 m/s, m₂ = 1500 kg, u₂ = 0
Common velocity after collision:
\[ V = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \]
\[ = \frac{1000 \times 20 + 1500 \times 0}{1000 + 1500} \]
\[ = \frac{20000}{2500} \]
\[ = 8 \, \text{m/s} \]
Initial kinetic energy:
\[ K_i = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \]
\[ = \frac{1}{2} \times 1000 \times (20)^2 + 0 \]
\[ = 200000 \, \text{J} \]
Final kinetic energy:
\[ K_f = \frac{1}{2} (m_1 + m_2) V^2 \]
\[ = \frac{1}{2} \times 2500 \times (8)^2 \]
\[ = 80000 \, \text{J} \]
Energy lost:
\[ \Delta K = K_i - K_f \]
\[ = 200000 - 80000 \]
\[ = 120000 \, \text{J} \]
Answer: Common velocity = 8 m/s, Energy lost = 120,000 J
Problem 3: Impulse Calculation

A ball of mass 0.5 kg moving with velocity 10 m/s is hit by a bat. After the hit, the ball moves with velocity 15 m/s in the opposite direction. Calculate the impulse imparted to the ball.

Given: m = 0.5 kg, u = 10 m/s, v = -15 m/s (opposite direction)
Using impulse-momentum theorem:
\[ J = m v - m u \]
\[ = 0.5 \times (-15) - 0.5 \times 10 \]
\[ = -7.5 - 5 \]
\[ = -12.5 \, \text{N·s} \]
The negative sign indicates the impulse is in the direction opposite to the initial motion.
Magnitude of impulse = 12.5 N·s

Frequently Asked Questions

❓ What is the difference between elastic and inelastic collisions?

The key difference is in the conservation of kinetic energy:

  • Elastic Collision: Both momentum and kinetic energy are conserved. No energy is lost as heat, sound, or deformation.
  • Inelastic Collision: Momentum is conserved but kinetic energy is not conserved. Some kinetic energy is converted to other forms.
  • Completely Inelastic Collision: A special case where the objects stick together after collision, resulting in maximum kinetic energy loss.

In real-world scenarios, most collisions are inelastic to some degree, with perfectly elastic collisions being an idealization.

❓ Why is momentum always conserved in collisions?

Momentum conservation in collisions follows from Newton's third law and the impulse-momentum theorem:

  • During a collision, each object exerts an equal and opposite force on the other
  • These internal forces cancel out when considering the system as a whole
  • If external forces are negligible compared to collision forces, the total momentum remains constant
  • This holds true regardless of whether the collision is elastic or inelastic

Momentum conservation is one of the most fundamental principles in physics and applies to all isolated systems.

❓ How do I solve two-dimensional collision problems?

For two-dimensional collisions, follow these steps:

  1. Establish a coordinate system (typically x and y axes)
  2. Apply conservation of momentum separately in x and y directions
  3. For elastic collisions, also apply conservation of kinetic energy
  4. Use trigonometric relationships for angles
  5. Solve the system of equations for unknown quantities

For completely inelastic collisions in 2D, the objects stick together and move with a common velocity, which can be found using vector addition of momenta.

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