What is a collision? Momentum is always conserved
A collision is a short, strong interaction between bodies (they may not even touch, like charged particles). During the very short hit, internal forces are huge and outside forces can be ignored. By Newton's third law the forces on the two bodies are equal and opposite, so total momentum is conserved: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
Total energy is also conserved, but kinetic energy may not be.
Elastic and inelastic collisions
- Elastic: momentum and kinetic energy both conserved. Examples: hard steel balls, atoms and molecules in a gas (nearly).
- Inelastic: momentum conserved, some K lost (heat, sound, deformation). Most real collisions.
- Perfectly inelastic: bodies stick together and move with one common velocity; the largest loss of K.
Coefficient of restitution e = (speed of separation)/(speed of approach) = (v₂ − v₁)/(u₁ − u₂). e = 1 elastic, 0 < e < 1 inelastic, e = 0 perfectly inelastic.
Elastic collision in 1D: derivation
Body 2 at rest (u₂ = 0).
- Momentum: m₁u₁ = m₁v₁ + m₂v₂ → m₁(u₁ − v₁) = m₂v₂ … (1)
- Kinetic energy: ½m₁u₁² = ½m₁v₁² + ½m₂v₂² → m₁(u₁ − v₁)(u₁ + v₁) = m₂v₂² … (2)
- Divide (2) by (1): u₁ + v₁ = v₂. (Speed of separation = speed of approach, so e = 1.)
- Put v₂ in (1): v₁ = (m₁ − m₂)u₁/(m₁ + m₂), v₂ = 2m₁u₁/(m₁ + m₂).
Special cases: m₁ = m₂ → v₁ = 0, v₂ = u₁ (swap). m₁ ≪ m₂ (ball hits wall) → v₁ ≈ −u₁ (bounces back). m₁ ≫ m₂ → v₂ ≈ 2u₁.
Perfectly inelastic collision in 1D
They move together: m₁u₁ + m₂u₂ = (m₁ + m₂)v, so v = (m₁u₁ + m₂u₂)/(m₁ + m₂).
Loss of KE (u₂ = 0): ΔK = ½ · m₁m₂/(m₁ + m₂) · u₁². Example: 2 kg at 3 m/s hits 1 kg at rest: v = 2 m/s, ΔK = ½ × (2/3) × 9 = 3 J.
General e: v₁ = (m₁ − e m₂)u₁/(m₁ + m₂), v₂ = (1 + e)m₁u₁/(m₁ + m₂) (for u₂ = 0).
Collisions in two dimensions
Body 1 moves along x with u₁ and hits body 2 at rest. After the hit they move at angles θ₁ and θ₂ on either side of the x-axis.
- x: m₁u₁ = m₁v₁ cos θ₁ + m₂v₂ cos θ₂
- y: 0 = m₁v₁ sin θ₁ − m₂v₂ sin θ₂
- If elastic, also: ½m₁u₁² = ½m₁v₁² + ½m₂v₂².
There are 4 unknowns (v₁, v₂, θ₁, θ₂) but only 3 equations, so one of them must be measured (like one angle).
Equal masses, elastic, one at rest: θ₁ + θ₂ = 90°. Proof: momentum gives u⃗ = v⃗₁ + v⃗₂; energy gives u² = v₁² + v₂². Squaring the first: u² = v₁² + v₂² + 2v⃗₁·v⃗₂, so v⃗₁·v⃗₂ = 0: the velocities are perpendicular.
Try it at home
Line up two same-size coins on a smooth table. Flick one straight into the other: the first nearly stops, the second moves on. Now flick it slightly off-centre and mark where each goes: the paths make almost a right angle. Then drop a rubber ball and a lump of clay from the same height: the ball bounces (e close to 1), the clay stays (e ≈ 0).
Exam corner
CBSE often asks: derive v₁ and v₂ for a 1D elastic collision (5 marks), show equal masses exchange velocities, find KE lost in a perfectly inelastic collision, and prove the 90° result in 2D. Always write momentum first, then energy.
Key formulas and definitions
- m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (always)
- e = (v₂ − v₁)/(u₁ − u₂)
- Elastic 1D (u₂ = 0): v₁ = (m₁ − m₂)u₁/(m₁ + m₂), v₂ = 2m₁u₁/(m₁ + m₂)
- Perfectly inelastic: v = (m₁u₁ + m₂u₂)/(m₁ + m₂)
- KE lost (u₂ = 0): ΔK = ½ · m₁m₂/(m₁ + m₂) · u₁²
- 2D: momentum conserved along x and y separately; equal masses elastic → θ₁ + θ₂ = 90°
Worked examples
1. A 3 kg ball at 4 m/s hits a 1 kg ball at rest elastically. Find both final velocities.
Step 1: v₁ = (3 − 1) × 4 / 4 = 2 m/s. Step 2: v₂ = 2 × 3 × 4 / 4 = 6 m/s. Check p: 12 = 6 + 6 ✔. Check K: 24 = 6 + 18 ✔.
2. A 1 kg ball at 6 m/s hits a 2 kg ball at rest elastically. Find the final velocities.
Step 1: v₁ = (1 − 2) × 6 / 3 = −2 m/s (bounces back). Step 2: v₂ = 2 × 1 × 6 / 3 = 4 m/s.
3. A 5 kg trolley at 4 m/s hits a 3 kg trolley at rest and they stick. Find the common speed and the KE lost.
Step 1: v = 5 × 4 / 8 = 2.5 m/s. Step 2: K before = ½ × 5 × 16 = 40 J. Step 3: K after = ½ × 8 × 6.25 = 25 J. Answer: 15 J lost.
4. A 10 g bullet at 400 m/s gets stuck in a 2 kg wooden block hanging at rest. Find the speed of the block just after.
Step 1: perfectly inelastic. Step 2: v = 0.01 × 400 / 2.01 ≈ 1.99 m/s.
5. For the block above, how high does it swing?
Step 1: after the hit, energy is conserved: ½v² = gh. Step 2: h = (1.99)² / 20 ≈ 0.198 m ≈ 20 cm. (This is the ballistic pendulum.)
6. A ball dropped from 1.6 m on a floor with e = 0.5. How high does it rise?
Step 1: speed at floor = √(2 × 10 × 1.6) = √32. Step 2: after bounce, v = 0.5 × √32. Step 3: h = v²/2g = 0.25 × 32 / 20 = 0.4 m (h′ = e²h).
7. Two equal pucks: A at 4 m/s hits B at rest elastically; A goes off at 30° to its old line. Find θ₂, v₁ and v₂.
Step 1: equal masses, elastic → θ₁ + θ₂ = 90°, so θ₂ = 60°. Step 2: x: 4 = v₁ cos30° + v₂ cos60°. y: v₁ sin30° = v₂ sin60°. Step 3: v₁ = 4 cos30° ≈ 3.46 m/s, v₂ = 4 cos60° = 2 m/s. Check K: 16 = 12 + 4 ✔.
8. A 2 kg ball moving east at 3 m/s hits a 1 kg ball moving north at 6 m/s and they stick. Find the common velocity.
Step 1: pₓ = 2 × 3 = 6, p_y = 1 × 6 = 6 (kg m/s). Step 2: p = √(36 + 36) = 6√2. Step 3: v = 6√2 / 3 = 2√2 ≈ 2.83 m/s at 45° north of east.
Common mistakes
- Using KE conservation in every collision. Only elastic collisions conserve kinetic energy; momentum is always conserved.
- Forgetting signs. Velocity to the left is negative; a negative v₁ means the ball bounces back.
- Adding 2D momenta as plain numbers. Split into x and y parts and conserve each separately.
- Thinking energy is "destroyed" in an inelastic collision. It becomes heat, sound and deformation; total energy is conserved.