Conservation of Momentum Formula: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

The conservation of momentum formula for two bodies is $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ : the total momentum before they interact equals the total momentum after, provided no net external force acts on the pair. It holds for gentle bumps, violent crashes and explosions alike.

The law follows from Newton's third law. During a collision each body pushes the other with an equal and opposite force for the same time, so the momentum one gains the other loses. That makes it one of the most useful results of the laws of motion. Give each velocity a sign when you use the calculator below; for bodies that stick together, enter the same value for v₁ and v₂.

Quick facts

Formula
$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$
Bodies that stick
$m_1u_1 + m_2u_2 = (m_1 + m_2)v$
From rest (explosion)
$0 = m_1v_1 + m_2v_2$
Condition
No net external force on the system
Unit of momentum
kg m/s
Kinetic energy
Conserved only in elastic collisions

Why the momentum conservation formula works

Let body 1 push on body 2 with force F for a short time Δt. By the third law, body 2 pushes back on body 1 with −F for the same Δt. The impulses are $+F\Delta t$ on body 2 and $-F\Delta t$ on body 1, so

$(m_2v_2 - m_2u_2) = -(m_1v_1 - m_1u_1)$

Rearranging gives the formula at the top. The internal forces cancel in pairs; only an outside force such as friction from the ground could change the total. In a quick collision outside forces act for so short a time that you can ignore them.

Using the conservation of momentum formula in five steps

Define the system: the bodies whose momenta you will add.
Check that external forces are absent, balanced, or too brief to matter.
Choose a positive direction and give every velocity a sign.
Write total momentum before = total momentum after, then solve for the unknown.
If asked, compare kinetic energies to decide whether the collision was elastic.

Solved examples on the formula for conservation of momentum

Example 1: trolleys that stick

A 2.0 kg trolley moving at 3.0 m/s hits a 1.0 kg trolley at rest. They stick together. Find their common velocity.

$2.0 \times 3.0 + 1.0 \times 0 = (2.0 + 1.0)v \Rightarrow v = \frac{6.0}{3.0} = 2.0\ \text{m/s}$

Kinetic energy before: ½ × 2.0 × 3.0² = 9.0 J. After: ½ × 3.0 × 2.0² = 6.0 J. One third of the kinetic energy became heat and sound, yet momentum was fully conserved.

Example 2: skaters pushing apart

Two skaters, 50 kg and 70 kg, stand at rest on smooth ice and push off each other. The 70 kg skater moves away at 2.0 m/s. Find the other skater's velocity.

Total momentum before is zero: $0 = 70 \times 2.0 + 50 \times v$ .

$v = -\frac{140}{50} = -2.8\ \text{m/s}$

The lighter skater moves at 2.8 m/s in the opposite direction.

Example 3: railway wagons, with km/h

A 20 000 kg wagon rolling at 36 km/h couples with a 30 000 kg wagon at rest. Find the speed after coupling in km/h.

Convert: 36 ÷ 3.6 = 10 m/s.

$v = \frac{20\,000 \times 10}{50\,000} = 4.0\ \text{m/s} = 4.0 \times 3.6 = 14.4\ \text{km/h}$

Elastic and inelastic collisions

Types of collision
Type	Momentum	Kinetic energy	Example
Elastic	Conserved	Conserved	Gas molecules, near enough for hard steel balls
Inelastic	Conserved	Some lost to heat, sound, deformation	A football hitting a wall
Perfectly inelastic	Conserved	Largest possible loss; bodies move together	Trolleys with velcro, Example 1

Momentum is conserved in every type. What changes is how much kinetic energy survives, measured by the coefficient of restitution. For a one dimensional elastic collision with body 2 at rest, $v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1$ and $v_2 = \frac{2m_1}{m_1 + m_2}u_1$ . Equal masses simply swap velocities.

Explosions and recoil

If the system starts at rest, the total momentum is zero and stays zero. Pieces fly apart with momenta that cancel: $m_1v_1 = -m_2v_2$ . A gun and its bullet, a cannon and its shell, and a person jumping off a small boat all follow this. The recoil velocity formula page applies it in detail, and rockets carry the same idea into continuous form on the thrust formula page.

Pitfalls with conservation of momentum

Leaving out signs, so that bodies moving toward each other seem to add momentum.
Applying conservation when a large outside force acts for a long time, such as friction during a long slide after the crash.
Assuming kinetic energy is conserved too. Only check that if the collision is stated to be elastic.
Using the formula on one body alone. It applies to the system.

In two dimensions, conserve the x momentum and the y momentum separately. That gives two equations.

Background on the quantity itself, with single body examples, sits under the momentum formula.

Frequently asked questions

What is the formula for conservation of momentum?

For two bodies, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. The total momentum before equals the total after when no net external force acts.

When is momentum not conserved?

When a net external force acts on the system, such as friction from the ground over a long time or a push from outside. Internal forces never change the total.

Is kinetic energy conserved when momentum is conserved?

Not necessarily. Only an elastic collision keeps the total kinetic energy unchanged, while momentum survives every collision of an isolated system.

How does conservation of momentum follow from Newton's laws?

The two bodies exert equal and opposite forces on each other for the same time, by the third law. Their impulses are equal and opposite, so the changes in momentum cancel.