Elastic and inelastic collisions in one and two dimensions
Momentum is conserved in every collision. Kinetic energy is conserved only in an elastic collision; in a perfectly inelastic one the bodies stick together.
What a collision is
When two bodies hit each other, they push on each other with large forces for a very short time. This is a collision. Outside forces, such as friction, are small during that time, so the pair forms an isolated system. So momentum is conserved in every collision. In a head-on (central) collision, the bodies move along the line joining their centres, before and after. A collision is perfectly elastic if the forces between the bodies are conservative, so the total kinetic energy is also conserved. It is perfectly inelastic if the bodies stick together and move as one. Most real collisions lie in between. Here, body 1 has mass m₁, velocity u₁ before and v₁ after. Your NIOS lesson writes these as mA, vAi and vAf, and as mB, vBi and vBf for body 2.
Elastic and inelastic collisions compared
| Feature | Elastic collision | Perfectly inelastic collision |
|---|---|---|
| Momentum | Conserved | Conserved |
| Kinetic energy | Conserved | Not conserved; part becomes heat, sound and change of shape |
| After the collision | The bodies separate | The bodies move together with one common velocity |
| Forces between the bodies | Conservative | Non-conservative |
| Examples | Collisions of gas molecules; very nearly, hard steel or glass balls and billiard balls | A bullet that gets stuck in a wooden block; a ball of wet clay sticking to a wall |
Elastic head-on collision: the equations and their results
- 1Body 1 (mass m₁, velocity u₁) hits body 2 (mass m₂, velocity u₂) head-on. After the collision their velocities are v₁ and v₂.
- 2Momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
- 3Kinetic energy: ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂².
- 4Solving the two together gives v₂ − v₁ = u₁ − u₂: the bodies separate as fast as they approached.
- 5The final velocities are v₁ = [(m₁ − m₂)u₁ + 2m₂u₂]/(m₁ + m₂) and v₂ = [(m₂ − m₁)u₂ + 2m₁u₁]/(m₁ + m₂).Some copies of the NIOS lesson print a minus sign before 2m₁u₁ (written 2mAvAi there) in the formula for v₂. That is a misprint: the plus sign is correct, as the lesson's own result for equal masses (v₂ = u₁) shows.
- 6★So, for equal masses, v₁ = u₂ and v₂ = u₁: the bodies exchange velocities. If body 2 was at rest, body 1 stops and body 2 moves off with u₁. If body 2 is very heavy and at rest, body 1 bounces back with −u₁ and body 2 stays still, like a ball hitting a wall.
★ key step
Worked example: an elastic head-on collision
A 1 kg ball moving at 6 m s⁻¹ hits a 2 kg ball at rest, head-on, in an elastic collision. Find their velocities after the collision.
- Here m₁ = 1 kg, u₁ = 6 m s⁻¹, m₂ = 2 kg, u₂ = 0.
- v₁ = (m₁ − m₂)u₁/(m₁ + m₂) = (1 − 2) × 6 ÷ 3 = −2 m s⁻¹.
- v₂ = 2m₁u₁/(m₁ + m₂) = 2 × 1 × 6 ÷ 3 = 4 m s⁻¹.
- Check momentum: before 1 × 6 = 6; after 1 × (−2) + 2 × 4 = 6 kg m s⁻¹.
- Check kinetic energy: before ½ × 1 × 36 = 18 J; after ½ × 1 × 4 + ½ × 2 × 16 = 2 + 16 = 18 J.
The 1 kg ball bounces back at 2 m s⁻¹; the 2 kg ball moves forward at 4 m s⁻¹
Worked example: a bullet stuck in a block (perfectly inelastic)
A 20 g bullet moving at 300 m s⁻¹ hits a 2 kg wooden block at rest on a smooth floor and stays inside it. Find the common velocity and the kinetic energy lost.
- Momentum before = 0.020 kg × 300 m s⁻¹ = 6 kg m s⁻¹.
- After: (0.020 + 2) kg × v = 6, so v = 6 ÷ 2.02 ≈ 2.97 m s⁻¹.
- Kinetic energy before = ½ × 0.020 × 300² = 900 J.
- Kinetic energy after = ½ × 2.02 × (2.97)² ≈ 8.9 J.
- Kinetic energy lost ≈ 900 − 8.9 ≈ 891 J, which becomes heat, sound and damage to the wood.
Common velocity about 2.97 m s⁻¹; about 891 J (99 percent) of the kinetic energy is lost
Extra, beyond your NIOS lesson: a collision in two dimensions
On a billiards table, a ball moving at 4 m s⁻¹ along the x-axis strikes an identical ball at rest, off-centre. After the collision, the first ball moves at 3.46 m s⁻¹ at 30° above the x-axis. Find the speed and direction of the second ball, ignoring friction and spin. Is the collision elastic?
- Your NIOS lesson works only with head-on collisions. This example shows that the same momentum rule also works in two dimensions.
- When bodies do not hit head-on, they fly off at angles. Momentum is a vector, so it is conserved separately along x and along y. Let each ball have mass m.
- Along x: m × 4 = m × 3.46 cos 30° + m v₂ cos θ₂. Since 3.46 × 0.866 ≈ 3.0, v₂ cos θ₂ = 4 − 3.0 = 1.0 m s⁻¹.
- Along y: 0 = m × 3.46 sin 30° − m v₂ sin θ₂, so v₂ sin θ₂ = 3.46 × 0.5 = 1.73 m s⁻¹.
- v₂ = √(1.0² + 1.73²) = √4.0 = 2.0 m s⁻¹, and tan θ₂ = 1.73, so θ₂ = 60° below the x-axis.
- Kinetic energy before = ½m × 4² = 8m; after = ½m × 3.46² + ½m × 2.0² = 6m + 2m = 8m. They are equal.
The second ball moves at 2.0 m s⁻¹ at 60° below the x-axis, and the collision is elastic. The two balls move off at 30° + 60° = 90° to each other, as always happens when equal masses collide elastically off-centre with one at rest
In a nuclear power plant, fast neutrons are slowed down by colliding with light nuclei in a moderator, such as heavy water. Carrom and billiards players use these collision rules in every game.
2 marks: “Distinguish between elastic and perfectly inelastic collisions.” Momentum is conserved in both. Elastic: kinetic energy is also conserved and the bodies separate. Perfectly inelastic: the bodies stick together and move with one velocity; part of the kinetic energy becomes heat, sound and deformation.
5 marks: “Two balls of masses m₁ and m₂, moving with velocities u₁ and u₂, collide head-on elastically. Write the equations of conservation of momentum and of kinetic energy, give the final velocities, and show that equal masses exchange velocities.” m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂; ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂². Results: v₁ = [(m₁ − m₂)u₁ + 2m₂u₂]/(m₁ + m₂), v₂ = [(m₂ − m₁)u₂ + 2m₁u₁]/(m₁ + m₂). Put m₁ = m₂: v₁ = u₂ and v₂ = u₁.
3 marks: “A 10 g bullet moving at 400 m s⁻¹ gets stuck in a 1.99 kg block at rest on a smooth surface. Find their common velocity and the loss of kinetic energy.” Momentum: 0.010 × 400 = 4 kg m s⁻¹ = 2.00 kg × v, so v = 2 m s⁻¹. KE before = ½ × 0.010 × 400² = 800 J; after = ½ × 2.00 × 2² = 4 J. Loss = 796 J.
Momentum: always conserved. Kinetic energy: only if elastic. Equal masses, elastic, head-on: swap velocities.