Conservation of momentum: collide two carts and watch

How to use this simulation

1. Run it with the defaults first: cart 1 moves right, cart 2 moves left, they collide midway, then move apart at their new velocities.
2. Drag either cart to change its velocity — the number updates instantly, and the bar charts below preview what the collision will do before you even press play.
3. Set e = 0: after the collision both carts move together at the same velocity, and the kinetic-energy bar drops sharply while the momentum bar stays the same height.
4. Set e = 1: the two kinetic-energy bars come out exactly the same height — no energy is lost at all.
5. Change m₁ or m₂ and run it again: check how v₁′ and v₂′ shift, using the same formula from the worked examples below.

Carrom, billiards and two train carriages coupling

Strike a carrom coin with the striker and the striker itself slows down or stops while the coin shoots off. A cue ball hitting a coloured ball on a billiards table does almost the same thing. Both raise the same question: can you predict exactly how fast each object moves after the impact, before it even happens?

You have probably watched two railway carriages coupling at a station — one rolls in slowly, clanks against the other, and then both carriages move on together at one shared speed. Here the two bodies do not bounce apart at all; they stay stuck together. That is a collision too, just a very different kind — a perfectly inelastic collision.

Between "bouncing apart cleanly" and "sticking together completely" there is an entire spectrum of possibilities, and a single number — the coefficient of restitution, e — tells you exactly where any real collision sits on that spectrum. Drag the e slider in the simulation above to see the whole range for yourself.

Starting from zero: momentum and its conservation

The product of an object's mass and velocity is called momentum, p = mv. Momentum is a vector — it has direction, so if motion to the right is taken as positive, motion to the left counts as negative. The total momentum of several objects is the vector sum of each one's momentum.

Newton's third law says that during a collision, the two bodies exert equal and opposite forces on each other, for exactly the same length of time. Whatever momentum one body gains, the other loses exactly that much, so the total stays unchanged. That is the conservation of momentum: when no net external force acts on a system, its total momentum stays constant.

Notice this law needs to know nothing about the details of the collision itself — how much the bodies deformed, how much heat was produced, none of it. Comparing the momentum just before and just after is enough. That is exactly what makes conservation of momentum one of physics's most powerful tools.

Kinetic energy (½mv²), on the other hand, is a scalar, and it is not automatically conserved in a collision — some can be lost to deformation, heat and sound. Exactly how much is lost depends on how "elastic" the collision is, which is measured by the coefficient of restitution.

Key terms in collisions

Get the vocabulary straight before the formulas; definition questions in exams come straight from this table.

Deriving the two formulas from the coefficient of restitution

Two carts of mass m₁, m₂ and initial velocities u₁, u₂ collide and end up moving at v₁′, v₂′ (whether they separate or move together). Two equations pin these down completely.

The first is conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁′ + m₂v₂′. The second is the definition of the coefficient of restitution: e = (v₂′ − v₁′) / (u₁ − u₂) — the speed at which they separate afterwards, as a fraction of the speed at which they approached beforehand. e = 1 means they separate at exactly the approach speed (elastic); e = 0 means the separation speed is zero, i.e. they move together.

Solving the two equations together gives two direct formulas that work for any value of e between 0 and 1 in one shot, with no separate cases to handle. Set e = 0 and both formulas collapse to the same shared velocity; set e = 1 and they become the familiar textbook elastic-collision formulas.

What the formulas say at e = 0 and e = 1

Setting e = 0 makes (m₁ − em₂) become m₁ and (m₂ − em₁) become m₂, and both formulas collapse to v₁′ = v₂′ = (m₁u₁ + m₂u₂)/(m₁ + m₂), the mass-weighted average velocity. That is exactly a perfectly inelastic collision: the two bodies move together at one shared speed.

Setting e = 1 turns the formulas into the classic elastic-collision pair: v₁′ = [(m₁−m₂)u₁ + 2m₂u₂]/(m₁+m₂), v₂′ = [2m₁u₁ + (m₂−m₁)u₂]/(m₁+m₂). This is exactly where you can see that two equal masses colliding elastically (one initially at rest) simply exchange velocities — Problem 3 below shows this directly.

Try these experiments in the simulation

Predict what will happen before each experiment, then check.

• Keep e = 0.5 and slowly raise cart 2's mass. Does v₂′ increase or decrease?
• Make the two masses equal, keep u₂ = 0, and set e = 1. Which cart moves after the collision?
• Set e = 0 and pick velocities so the total momentum is exactly zero (equal masses, equal and opposite velocities). Do both carts end up at rest after the collision?
• Keeping the same setup, raise e from 0 to 1 gradually. How does the "after" kinetic-energy bar grow?
• Drag cart 1 so it can never catch cart 2 (slower, or moving the other way). What message appears?

Solved problems

Every solution states what is given, then the formula, then the substitution — write it this way in exams to get full marks.

How kinetic energy comes back as the restitution rises

Holding the simulation's own default setup fixed (m₁ = 2 kg, u₁ = 4 m/s, m₂ = 3 kg, u₂ = -1 m/s) and changing only e, momentum stays exactly the same throughout, while kinetic energy rises steadily with e, fully returning at e = 1.

Common mistakes

Avoiding these keeps marks safe on both the numeric and the conceptual questions.

• Assuming both momentum and kinetic energy are always conserved. Momentum always is; kinetic energy only is when e = 1.
• Adding momenta as if they were scalars, ignoring direction. Forgetting to make an opposite-direction velocity negative gives the wrong answer.
• Assuming "elastic collision" means "a hard, fast impact". Elastic vs inelastic is about whether kinetic energy is conserved, not about how forceful the collision looks.
• Assuming e = 0 means both velocities become zero. It actually means the two bodies end up moving together, usually at some nonzero shared velocity.
• Swapping which mass goes where in the formula. Whichever cart's final velocity you are solving for, its own mass appears first in that formula.
• Getting the order of subtraction backwards in the restitution definition. e = (v₂′ − v₁′)/(u₁ − u₂); reversing either subtraction flips the sign.
• Assuming the coefficient of restitution can exceed 1 or be negative. For a real collision, 0 ≤ e ≤ 1 always.

Real-life uses of momentum conservation

Beyond the exam, conservation of momentum underlies many safety and engineering decisions.

• A car's crumple zone: the front of the car is deliberately designed to crush gradually, stretching out the collision time so the force on the passengers is smaller.
• Rocket launches: a rocket pushes exhaust gas backward to move itself forward — the same conservation of momentum at work as a recoiling gun.
• Billiards and carrom: nearly elastic collisions, which is exactly why players can calculate speed and angle to predict where a ball or coin will end up.
• Traffic-accident reconstruction: skid marks and the final resting positions of vehicles let investigators use conservation of momentum to work backward to the speeds before impact.
• Spacecraft docking: bringing two spacecraft together is close to a perfectly inelastic collision, so their velocities must be matched very carefully first.
• Newton's cradle: a chain of nearly elastic collisions that makes both momentum and kinetic energy conservation visible at the same time.

Exam corner

Conservation of momentum and collisions is a staple topic in school and college physics: expect a statement-and-derive question on the conservation law, a conceptual question distinguishing elastic from inelastic collisions, and a numeric question using the restitution formulas.

Revision: the one-screen summary

This list and the table above are all you need the night before an exam.

• Momentum p = mv is a vector. With no outside force, total momentum is always conserved.
• Coefficient of restitution e = (v₂′−v₁′)/(u₁−u₂); e = 1 is elastic, e = 0 is perfectly inelastic.
• General formulas: v₁′ = [(m₁−em₂)u₁+(1+e)m₂u₂]/(m₁+m₂), v₂′ = [(1+e)m₁u₁+(m₂−em₁)u₂]/(m₁+m₂).
• At e = 0, v₁′ = v₂′ = (m₁u₁+m₂u₂)/(m₁+m₂); at e = 1, kinetic energy is conserved too.
• Equal masses in an elastic collision (one initially at rest) simply exchange velocities.
• Kinetic energy is conserved only in an elastic collision; an inelastic one loses some to heat and sound, and a perfectly inelastic one loses the most.

Frequently asked questions