Momentum & Collisions
High SchoolMomentum and its conservation
Linear momentum is mass times velocity, a vector:
Newton's second law is truly ΣF = dp/dt. The impulse of a force is J = ∫F dt = Δp — a large force for a short time, or a small force for long, can give the same momentum change.
Conservation of momentum: if no external force acts on a system, its total momentum is constant. This follows directly from the third law — internal forces cancel in pairs. For a two-body collision:
Types of collision
Elastic: kinetic energy conserved and momentum conserved. Inelastic: momentum conserved but KE lost (to heat, deformation, sound). Perfectly inelastic: bodies stick together, maximum KE lost consistent with momentum conservation.
Why momentum is conserved when energy isn't
Momentum is always conserved in an isolated system, in every collision, whereas kinetic energy is conserved only in elastic ones. Why the asymmetry? Momentum conservation is a direct consequence of Newton's third law: whatever momentum one body gains, the other loses exactly. Kinetic energy, by contrast, can be converted into other forms (heat, sound, permanent deformation) that don't cancel out.
This is why impulse thinking saves lives. Airbags, crumple zones and catching a ball by drawing your hand back all increase the collision time Δt. Since Δp is fixed (you must stop), a longer Δt means a smaller force F = Δp/Δt — and it's force, not momentum change, that injures.
Worked example — a perfectly inelastic collision
A 1500 kg car at 20 m·s⁻¹ rear-ends a stationary 1000 kg car; they lock together. Find the common velocity afterward and the kinetic energy lost.
- Conserve momentum:
m₁u₁ = (m₁+m₂)v. v = (1500×20)/(1500+1000) = 30000/2500 = 12.0 m·s⁻¹.- KE before:
½×1500×20² = 300 000 J. - KE after:
½×2500×12² = 180 000 J. - KE lost:
300 000 − 180 000 = 120 000 J(40%), converted to heat, sound and crumpling.
Momentum is fully conserved (300 000 kg·m·s⁻¹ before and after), yet 40% of the kinetic energy is gone — exactly what defines an inelastic collision.
A 0.15 kg ball hits a wall at 12 m·s⁻¹ and rebounds at 8 m·s⁻¹. Find the impulse on the ball.
Take toward the wall as positive. Δp = m(v − u) = 0.15(−8 − 12) = 0.15×(−20) = −3.0 kg·m·s⁻¹. The impulse is 3.0 N·s directed away from the wall. Note we must use the sign change on rebound — that's why it's larger than for a ball that merely stops.
In an elastic head-on collision, a moving ball strikes an identical stationary ball. What happens?
The moving ball stops dead and the target moves off with the incoming speed. Solving momentum and energy conservation for equal masses forces a complete exchange of velocities — the familiar Newton's-cradle result.
Why does a gun recoil?
Before firing, total momentum is zero. Afterward the bullet carries momentum forward, so the gun must carry equal momentum backward: mgunvgun = mbulletvbullet. Because the gun is far heavier, its recoil speed is small — but the momenta are equal and opposite.
How do airbags reduce injury if you still lose the same momentum?
They extend the stopping time Δt. Momentum change Δp is fixed by your mass and speed, but average force is F = Δp/Δt. Increasing Δt tenfold cuts the peak force tenfold — and force is what breaks bones.
How the ideas connect
Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.
The key facts, visualised
Worked problems, step by step
Follow each solution line by line, then try to reproduce it on paper before moving on.
Example 1A 2.0 kg cart at 3.0 m/s hits and sticks to a 1.0 kg cart at rest. Find their common speed.
- Momentum before = 2.0*3.0 + 1.0*0 = 6.0 kg*m/s
- After: (2.0 + 1.0)*v = 6.0
- v = 6.0 / 3.0
Example 2A 0.5 kg ball hits a wall at 4 m/s and bounces back at 4 m/s in 0.02 s. Find the average force.
- Change in momentum = 0.5*(4 - (-4)) = 4.0 kg*m/s
- F = impulse / time = 4.0 / 0.02
Now you try
Work each one out first, then tap to reveal the worked answer.