Conservation of Linear Momentum in Isolated Systems

The law of conservation of linear momentum states that the total momentum of an isolated system (one with no net external force) remains constant: the total momentum before an interaction equals the total momentum after it. In brief: for two bodies, m1u1 + m2u2 = m1v1 + m2v2. The law follows from Newton’s third law: internal forces act in equal and opposite pairs, so they cannot change the total momentum. It applies to collisions, explosions and recoil, whether or not kinetic energy is conserved.

Billiard balls, railcars coupling and rockets all obey this law. It is often easier to use than Newton’s second law, because the details of the forces during the interaction do not matter. This guide states the law and its conditions, gives a method for problems, compares types of interactions in a table, works through numerical examples and corrects common mistakes.

What does the law say?

If no net external force acts on a system, its total momentum ptotal = Σmv does not change. An isolated system has no external forces, or the external forces cancel. Often the law is applied approximately to a short collision where the internal forces are much larger than external ones such as friction (the system is nearly isolated for that short time). Momentum is conserved separately in each direction; if an external force acts only in one direction (gravity downward), momentum is conserved in the perpendicular directions.

Why is momentum conserved?

In a collision between A and B, A exerts a force F on B for a time Δt, and by the third law, B exerts −F on A for the same time. The impulses are equal and opposite: ΔpB = FΔt and ΔpA = −FΔt. Therefore, ΔpA + ΔpB = 0, and the total does not change. Momentum is transferred from one object to the other, but not created or destroyed. The law is linked to a symmetry of nature, the homogeneity of space (Noether’s theorem), and it holds in relativity and quantum mechanics too.

How is the law used?

  1. Define the system so that it is isolated, and choose a positive direction.
  2. Write the total momentum before the interaction, with signs.
  3. Write the total momentum after.
  4. Set them equal and solve for the unknown velocity or mass.
  5. Check the energy to find the type of collision, if needed.

Types of interactions compared

Conservation in different interactions
Interaction Momentum conserved? Kinetic energy conserved?
Elastic collisionYesYes
Inelastic collisionYesNo (some lost)
Perfectly inelastic (stick together)YesNo (maximum loss)
Explosion or recoilYesNo (increases)
Two railcars coupling together, before and after Before coupling 2,000 kg 3,000 kg 3.0 m/s at rest After coupling 5,000 kg 1.2 m/s Momentum: 6,000 kg·m/s before and after Kinetic energy: 9,000 J → 3,600 J
The coupling example. Total momentum is the same before and after (2,000 × 3.0 = 5,000 × 1.2 = 6,000 kg·m/s), so the combined cars move at 1.2 m/s. Arrow lengths are proportional to speed. The kinetic energy falls from 9,000 J to 3,600 J, because the collision is inelastic.

Worked example: coupling railcars

  1. Coupling. A railcar of mass 2,000 kg moving at 3.0 m/s collides with a stationary car of mass 3,000 kg, and the two couple together. Momentum before = 2,000 × 3.0 + 3,000 × 0 = 6,000 kg·m/s. After, the mass is 5,000 kg: v = 6,000 ÷ 5,000 = 1.2 m/s.
  2. Energy check. Kinetic energy before = ½ × 2,000 × 3.02 = 9,000 J. After = ½ × 5,000 × 1.22 = 3,600 J. The energy lost is 5,400 J (60 percent), converted to thermal energy, sound and deformation. This is an inelastic collision.
  3. Car crash. A 1,000 kg car at 15 m/s hits the rear of a 1,500 kg car moving at 5.0 m/s in the same direction, and they stick together. p = 1,000 × 15 + 1,500 × 5.0 = 15,000 + 7,500 = 22,500. v = 22,500 ÷ 2,500 = 9.0 m/s.
  4. Head-on. A 60 kg skater moving at +4.0 m/s grabs a 40 kg skater moving at −2.0 m/s. p = 240 − 80 = 160 kg·m/s, v = 160 ÷ 100 = +1.6 m/s, in the direction of the first skater.
  5. Catching a ball. A 70 kg person on ice catches a 2.0 kg ball thrown at 10 m/s. v = 2.0 × 10 ÷ 72 = 0.28 m/s.
2,000 kg 3,000 kg 3.0 m/s v = 0 p = 6,000 kg·m/s 5,000 kg 1.2 m/s p = 6,000 kg·m/s

How was the law discovered?

Descartes proposed conservation of “quantity of motion” in 1644. John Wallis, Christopher Wren and Christiaan Huygens corrected it for the direction in the 1660s, using experiments on collisions of pendulum bobs presented to the Royal Society. Newton derived it from his third law in the Principia. Emmy Noether showed in 1918 that it follows from the translational symmetry of space.

Which mistakes are common?

Mistake 1: Applying conservation when external forces act

Check that the system is isolated, or that the interaction is short enough for external impulses to be negligible.

Mistake 2: Forgetting the signs

Velocities in opposite directions have opposite signs.

Mistake 3: Assuming kinetic energy is conserved

Momentum is always conserved in isolated systems. Kinetic energy is conserved only in elastic collisions.

Mistake 4: Applying the law to a single object

The conserved quantity is the total momentum of all objects in the system. A single object’s momentum can change.