Prerequisites
Before studying this topic, review:
- Kinetic Energy — KE formula and its properties
- Conservation of Energy — When and how energy is conserved
- Newton’s Laws — Momentum and impulse concepts
What is a Collision?
A collision is a brief, strong interaction between two or more bodies. During collision:
- Large forces act for short time
- External forces (like gravity) can be neglected
- Momentum is always conserved
Interactive: Compare Collision Types
Click the buttons to see how elastic, inelastic, and perfectly inelastic collisions differ:
Interactive Collision Simulator
Explore collisions in depth with this full-featured simulator. Adjust masses, velocities, and collision types to see:
- How momentum is always conserved (watch the momentum arrows)
- How kinetic energy behaves differently for elastic vs inelastic collisions
- The effect of the coefficient of restitution
- Special cases like equal mass velocity exchange
- Equal masses: Set m1 = m2 = 5 kg, u1 = 5 m/s, u2 = 0 (elastic) - velocities swap!
- Heavy hits light: m1 = 10 kg, m2 = 1 kg, u1 = 5 m/s, u2 = 0 - watch the light ball fly!
- Perfectly inelastic: Any masses moving toward each other - see maximum KE loss
- Partial inelastic: Use the coefficient slider to explore intermediate cases
Types of Collisions
| Type | Momentum | Kinetic Energy | $e$ |
|---|---|---|---|
| Elastic | Conserved | Conserved | 1 |
| Inelastic | Conserved | Not conserved | 0 < e < 1 |
| Perfectly Inelastic | Conserved | Maximum loss | 0 |
Coefficient of Restitution
The coefficient of restitution $e$ measures the “bounciness” of a collision:
$$\boxed{e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{\text{velocity of separation}}{\text{velocity of approach}}}$$where:
- $u_1, u_2$ = initial velocities of bodies 1 and 2
- $v_1, v_2$ = final velocities of bodies 1 and 2
Values of e
| Collision Type | e Value | Physical Meaning |
|---|---|---|
| Perfectly elastic | $e = 1$ | No KE loss |
| Inelastic | $0 < e < 1$ | Partial KE loss |
| Perfectly inelastic | $e = 0$ | Bodies stick together |
Elastic Collision (1D)
For two bodies with masses $m_1$, $m_2$ and initial velocities $u_1$, $u_2$:
Conservation Equations
Momentum:
$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$Kinetic Energy:
$$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$$Final Velocity Formulas
$$\boxed{v_1 = \frac{(m_1 - m_2)u_1 + 2m_2u_2}{m_1 + m_2}}$$ $$\boxed{v_2 = \frac{(m_2 - m_1)u_2 + 2m_1u_1}{m_1 + m_2}}$$Special Cases of Elastic Collision
Case 1: Equal Masses ($m_1 = m_2$)
$$v_1 = u_2, \quad v_2 = u_1$$Velocities are exchanged!
If one body is at rest ($u_2 = 0$):
- Moving body stops ($v_1 = 0$)
- Stationary body moves with initial velocity ($v_2 = u_1$)
Case 2: Heavy Body Hits Light Body at Rest ($m_1 >> m_2$, $u_2 = 0$)
$$v_1 \approx u_1, \quad v_2 \approx 2u_1$$Heavy body continues almost unchanged; light body moves at twice the speed.
Case 3: Light Body Hits Heavy Body at Rest ($m_1 << m_2$, $u_2 = 0$)
$$v_1 \approx -u_1, \quad v_2 \approx 0$$Light body bounces back with nearly same speed; heavy body barely moves.
Case 4: Heavy Body at Rest ($m_2 >> m_1$, $u_2 = 0$)
$$v_1 \approx -u_1, \quad v_2 \approx 0$$Light body rebounds, heavy body remains stationary.
Perfectly Inelastic Collision
Bodies stick together after collision.
Final Velocity
$$\boxed{v = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}}$$This is the velocity of the combined mass $(m_1 + m_2)$.
Loss in Kinetic Energy
$$\boxed{\Delta KE = \frac{1}{2}\frac{m_1m_2}{m_1 + m_2}(u_1 - u_2)^2}$$This is the maximum possible loss for given masses and velocities.
Inelastic Collision (General)
Using coefficient of restitution $e$:
$$v_2 - v_1 = e(u_1 - u_2)$$Combined with momentum conservation:
$$v_1 = \frac{m_1u_1 + m_2u_2 - em_2(u_1 - u_2)}{m_1 + m_2}$$ $$v_2 = \frac{m_1u_1 + m_2u_2 + em_1(u_1 - u_2)}{m_1 + m_2}$$Loss in Kinetic Energy
$$\Delta KE = \frac{1}{2}\frac{m_1m_2}{m_1 + m_2}(1 - e^2)(u_1 - u_2)^2$$Two-Dimensional Collisions
For 2D collisions, apply conservation separately:
x-component:
$$m_1u_{1x} + m_2u_{2x} = m_1v_{1x} + m_2v_{2x}$$y-component:
$$m_1u_{1y} + m_2u_{2y} = m_1v_{1y} + m_2v_{2y}$$In a 2D elastic collision between equal masses where one is initially at rest, the bodies move perpendicular to each other after collision.
$$\vec{v_1} \perp \vec{v_2}$$Worked Examples
Example 1: Elastic Collision
Solution: Using formulas:
$$v_1 = \frac{(1-2)(3) + 2(2)(-2)}{1+2} = \frac{-3 - 8}{3} = \frac{-11}{3} = -3.67 \text{ m/s}$$ $$v_2 = \frac{(2-1)(-2) + 2(1)(3)}{1+2} = \frac{-2 + 6}{3} = \frac{4}{3} = 1.33 \text{ m/s}$$The 1 kg ball moves at 3.67 m/s (left) and 2 kg ball moves at 1.33 m/s (right).
Example 2: Perfectly Inelastic Collision
Solution:
Common velocity:
$$v = \frac{m_1u_1 + m_2u_2}{m_1 + m_2} = \frac{4(8) + 2(0)}{4 + 2} = \frac{32}{6} = \boxed{\frac{16}{3} \approx 5.33 \text{ m/s}}$$Energy loss:
$$\Delta KE = \frac{1}{2}\frac{m_1m_2}{m_1 + m_2}(u_1 - u_2)^2 = \frac{1}{2}\frac{4 \times 2}{6}(8)^2 = \frac{4}{6}(64) = \boxed{\frac{128}{3} \approx 42.67 \text{ J}}$$Example 3: Using Coefficient of Restitution
Solution: Wall is stationary before and after collision.
$$e = \frac{v_{ball} - v_{wall}}{u_{wall} - u_{ball}} = \frac{6 - 0}{0 - (-10)} = \frac{6}{10} = \boxed{0.6}$$Example 4: 2D Elastic Collision
Solution: For equal mass elastic collision, bodies move perpendicular after collision.
So the second ball moves at 90° - 30° = 60° to original direction.
Momentum conservation (x):
$$m(10) = m(v_1\cos30°) + m(v_2\cos60°)$$ $$10 = v_1(0.866) + v_2(0.5)$$Momentum conservation (y):
$$0 = m(v_1\sin30°) - m(v_2\sin60°)$$ $$v_1(0.5) = v_2(0.866)$$ $$v_1 = 1.732v_2$$Substituting:
$$10 = 1.732v_2(0.866) + v_2(0.5) = 1.5v_2 + 0.5v_2 = 2v_2$$ $$v_2 = \boxed{5 \text{ m/s}}, \quad v_1 = \boxed{8.66 \text{ m/s}}$$Summary Table
| Type | Final Velocities | KE Loss |
|---|---|---|
| Elastic | Use standard formulas | 0 |
| Perfectly inelastic | $v = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}$ | Maximum |
| Inelastic (general) | Use $e$ with momentum | $\propto (1-e^2)$ |
Practice Problems
A 3 kg ball moving at 4 m/s collides elastically with a 1 kg ball at rest. Find final velocities.
A ball dropped from height 2 m rebounds to height 1.28 m. Find coefficient of restitution.
Two balls (2 kg each) moving at 6 m/s toward each other collide inelastically ($e = 0.5$). Find final velocities and energy loss.
In a perfectly inelastic collision, show that the fraction of KE lost is $\frac{m_2}{m_1 + m_2}$ when $m_2$ is at rest.
Related Topics
Within Work, Energy & Power
- Work-Energy Theorem — Connects work and KE change
- Conservation of Energy — For elastic collisions
Connected Chapters
- Center of Mass — System approach to collisions
- Projectile Motion — 2D collision problems
- Relative Velocity — Velocity of approach/separation
Real-World Applications
- Atoms and Nuclei — Nuclear collisions and scattering
- Kinetic Theory — Molecular collisions in gases