Conservation of Momentum Calculator
Calculate final velocities and total momentum for elastic collisions, perfectly inelastic collisions, and general momentum conservation problems. Based on the law of conservation of momentum: the total momentum of an isolated system remains constant. Enter the masses and initial velocities of two objects to solve instantly.
The Law of Conservation of Momentum
The conservation of momentum law is one of the most fundamental principles in physics: the total momentum of an isolated system is constant if no net external force acts on it. For a two-body system, this means:
p_total = m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f
where m₁, m₂ are masses (kg), v₁ᵢ, v₂ᵢ are initial velocities (m/s), and v₁f, v₂f are final velocities after the interaction. The law applies to collisions, explosions, and any interaction between objects where external forces (friction, gravity components) are either absent or negligible during the brief interaction.
Conservation of momentum is a consequence of Newton's third law: action-reaction force pairs during a collision are equal and opposite, so the impulses they deliver to each object are equal and opposite, leaving total momentum unchanged. This derivation makes conservation of momentum more general than Newton's laws — it holds in special relativity and quantum mechanics too, as long as spatial symmetry (homogeneity of space) is preserved.
Elastic vs Inelastic Collisions — Formulas and Examples
Elastic collision formulas (momentum AND kinetic energy conserved):
v₁f = ((m₁−m₂)/(m₁+m₂))·v₁ᵢ + (2m₂/(m₁+m₂))·v₂ᵢ
v₂f = (2m₁/(m₁+m₂))·v₁ᵢ + ((m₂−m₁)/(m₁+m₂))·v₂ᵢ
Example (elastic): A 3 kg ball at 4 m/s hits a stationary 1 kg ball. v₁f = ((3−1)/4)×4 = 2 m/s; v₂f = (6/4)×4 = 6 m/s. Check KE: initial = ½×3×16 = 24 J; final = ½×3×4 + ½×1×36 = 6 + 18 = 24 J. Conserved.
Perfectly inelastic collision formula (objects stick; only momentum conserved):
vf = (m₁v₁ᵢ + m₂v₂ᵢ) / (m₁ + m₂)
Example (inelastic): A 2 kg clay ball at 6 m/s hits a 4 kg stationary ball and they stick. vf = (2×6 + 4×0) / 6 = 2 m/s. KE lost = ½×2×36 − ½×6×4 = 36 − 12 = 24 J converted to heat and deformation.
Real-World Applications of Conservation of Momentum
Vehicle crash analysis: Accident reconstructionists use momentum conservation to determine pre-crash speeds. If two cars collide and leave together (inelastic), measuring the final velocity and skid marks lets engineers calculate exactly how fast each car was traveling before impact — critical evidence in legal proceedings.
Rocket propulsion: A rocket expels exhaust gases backward at high velocity. The momentum gained by the exhaust equals the momentum gained by the rocket in the opposite direction. This is why rockets work in the vacuum of space — they don't need anything to push against, only the reaction of ejected mass.
Newton's cradle: The desk toy demonstrates elastic collision with equal masses: the incoming ball stops dead and the far ball swings out at the same speed, conserving both momentum and kinetic energy. If two balls swing in, two balls swing out — the number of moving balls encodes momentum and energy conservation simultaneously.
Particle physics: At CERN's Large Hadron Collider, detectors reconstruct particle collisions by tracking all fragments and verifying that total 4-momentum (relativistic generalization) is conserved. Missing momentum in the detector signals the creation of particles that escaped detection — the method used to infer the existence of neutrinos and dark matter candidates.
Frequently Asked Questions
What is the law of conservation of momentum?
The law of conservation of momentum states that the total momentum of an isolated system (no external forces) remains constant over time. For two objects: m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f. Momentum (p = mv) is a vector quantity — direction matters. If object 1 moves right (+) and object 2 moves left (−), assign negative velocity to object 2.
What is the difference between elastic and inelastic collisions?
In an elastic collision, both momentum AND kinetic energy are conserved. The objects bounce off each other with no energy loss to heat or deformation (e.g., ideal billiard balls, atomic collisions). In a perfectly inelastic collision, momentum is conserved but kinetic energy is not — the objects stick together and move as one. Real collisions fall between these extremes. The conservation of momentum equation applies to both types; the elastic case has an additional constraint (KE conserved) that fixes both final velocities.
How do you find final velocity after an elastic collision?
For a 1D elastic collision: v₁f = ((m₁−m₂)/(m₁+m₂))v₁ᵢ + (2m₂/(m₁+m₂))v₂ᵢ and v₂f = (2m₁/(m₁+m₂))v₁ᵢ + ((m₂−m₁)/(m₁+m₂))v₂ᵢ. Special case: if m₁ = m₂ and v₂ᵢ = 0, then v₁f = 0 and v₂f = v₁ᵢ — the first object stops and the second moves off at the original speed (familiar from Newton's cradle).
What is momentum in physics?
Momentum (p) is the product of an object's mass and velocity: p = mv. It is measured in kg·m/s. A 0.05 kg tennis ball at 50 m/s has momentum p = 0.05 × 50 = 2.5 kg·m/s. A 1000 kg car at 30 m/s has p = 30,000 kg·m/s — 12,000 times more. Momentum quantifies how hard it is to stop an object; heavier and faster objects have greater momentum.
How is conservation of momentum related to Newton's laws?
Conservation of momentum follows directly from Newton's third law. When two objects collide, they exert equal and opposite forces on each other (F₁₂ = −F₂₁). Over the same time interval, the impulse (force × time) each receives is equal and opposite, so the change in momentum of each is equal and opposite — total momentum is unchanged. More formally, if the net external force on a system is zero, then dp/dt = 0, meaning total momentum is constant.
Does conservation of momentum apply in explosions?
Yes. In an explosion, the system starts at rest (total momentum = 0) and internal forces push fragments apart. Since external forces are negligible during the explosion, total momentum remains zero — the fragments fly apart with momenta that vector-sum to zero. A 2 kg rocket expels 0.5 kg of gas at 600 m/s backward: gas momentum = −300 kg·m/s, so the rocket (1.5 kg) moves forward at 200 m/s to conserve the original zero total momentum.