Conservation of Momentum: Elastic & Inelastic Collisions | Shikkha Web

Q: What does the law of conservation of momentum state?

When no net external force acts on a system, its total momentum stays exactly the same before and after any collision or interaction.

Q: What is the difference between an elastic and an inelastic collision?

In an elastic collision (e = 1), both momentum and kinetic energy are conserved. In an inelastic collision (e < 1), momentum is still conserved, but some kinetic energy is lost to heat, sound or deformation.

Q: What is the coefficient of restitution?

The coefficient of restitution, e, is the ratio of the relative speed at which two bodies separate after a collision to the relative speed at which they approached before it: e = (v₂′−v₁′)/(u₁−u₂).

Q: What happens in a perfectly inelastic collision (e = 0)?

The two bodies do not separate after impact; they move together at one shared velocity, v₁′ = v₂′ = (m₁u₁+m₂u₂)/(m₁+m₂). This is where the largest possible fraction of kinetic energy is lost.

Q: What are the final velocities in the simulation's default collision (m₁ = 2 kg, u₁ = 4 m/s, m₂ = 3 kg, u₂ = -1 m/s, e = 0.50)?

Substituting into the formulas gives v₁′ = -0.50 m/s and v₂′ = 2.00 m/s. Momentum is 5 kg·m/s both before and after, while kinetic energy falls from 17.50 J to 6.25 J.

Q: What happens when two equal masses collide elastically?

If one is initially at rest, the moving one stops completely and the one that was at rest moves off with exactly the first one's original velocity — the velocities simply swap. This is what happens on a billiards table.

Q: Can the coefficient of restitution be found from velocities alone, without knowing the masses?

Yes. Measuring the velocities before and after a collision and substituting into e = (v₂′−v₁′)/(u₁−u₂) gives e directly; the masses never enter that formula.

Q: Where does the lost kinetic energy go in an inelastic collision?

It is not destroyed — just like momentum, energy is conserved overall. The kinetic energy "lost" from the bodies' motion is converted into heat, sound and permanent deformation of the colliding objects.

Q: Is a gun's recoil an example of conservation of momentum?

Yes. The total momentum was zero before firing, so it must still be zero afterward — if the bullet moves forward, the gun is forced to move backward by exactly the amount needed to keep the total at zero.

Drag either cart to set its velocity, or use the sliders.

Cart 1Cart 2BeforeAfter

Speed

Controls

Mass of cart 1, m₁2.0 kg

Mass of cart 2, m₂3.0 kg

Initial velocity of cart 1, u₁4.0 m/s

Initial velocity of cart 2, u₂-1.0 m/s

Coefficient of restitution, e (0 = stick, 1 = elastic)0.50

Readings

Cart 1's velocity right now: 4.00m/s
Cart 2's velocity right now: -1.00m/s
Cart 1's final velocity, v₁′: -0.50m/s
Cart 2's final velocity, v₂′: 2.00m/s
Total momentum (Before): 5.00kg·m/s
Total momentum (After): 5.00kg·m/s
Total kinetic energy (Before): 17.50J
Total kinetic energy (After): 6.25J
Kinetic energy lost: 11.25J

How to use this simulation

Run it with the defaults first: cart 1 moves right, cart 2 moves left, they collide midway, then move apart at their new velocities.
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.
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.
Set e = 1: the two kinetic-energy bars come out exactly the same height — no energy is lost at all.
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.

Term	Symbol	What it means	SI unit
Momentum	p = mv	A body's mass times its velocity, a vector	kg·m/s
Conservation of momentum	—	Total momentum stays constant with no external force	—
Elastic collision	e = 1	Both momentum and kinetic energy are conserved	—
Inelastic collision	e < 1	Momentum is conserved, but some kinetic energy is lost	—
Perfectly inelastic collision	e = 0	The two bodies stick together after impact	—
Coefficient of restitution	e	Ratio of separation speed after impact to approach speed before it	—
Kinetic energy	KE = ½mv²	Energy a body has because it is moving	joule (J)
Recoil	—	The backward motion of the remaining mass after an explosion or a shot	—

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.

e = (v₂′ − v₁′) / (u₁ − u₂)the definition of the coefficient of restitution

v₁′ = [(m₁ − em₂)u₁ + (1+e)m₂u₂] / (m₁ + m₂)both hold for any e from 0 to 1

v₂′ = [(1+e)m₁u₁ + (m₂ − em₁)u₂] / (m₁ + m₂)

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.

Problem 1: the simulation's default collision

m₁ = 2 kg, u₁ = 4 m/s (rightward) and m₂ = 3 kg, u₂ = -1 m/s (leftward), e = 0.50. What are the two final velocities?

v₁′ = [(m₁ − em₂)u₁ + (1+e)m₂u₂]/(m₁+m₂) = -0.50 m/s. v₂′ = [(1+e)m₁u₁ + (m₂−em₁)u₂]/(m₁+m₂) = 2.00 m/s. Check: p_before = m₁u₁+m₂u₂ = 5 kg·m/s, and m₁v₁′+m₂v₂′ also comes to 5 kg·m/s — momentum matches. Kinetic energy fell from 17.50 J to 6.25 J, a loss of 11.25 J.

Problem 2: a bullet embedding in a block (e = 0)

A 0.01 kg bullet moving at 400 m/s embeds itself in a stationary 3.99 kg wooden block. What speed do the bullet and block move at together?

With e = 0, v_common = (m_bullet u_bullet + m_block × 0)/(m_bullet + m_block) = (0.01 × 400)/4.00 = 1.00 m/s. The bullet's huge speed is shared across the whole combined mass, giving a much smaller final speed — this is exactly the principle behind a ballistic pendulum.

Problem 3: equal masses, elastic — velocities exchange

Two equal carts, each 2 kg. One moving at 5 m/s strikes the other, which is at rest (e = 1). What happens?

Substituting e = 1 and m₁ = m₂ gives v₁′ = 0 m/s (it stops) and v₂′ = 5 m/s (it takes on the first cart's entire velocity). In an elastic collision between equal masses, the velocities simply swap — this is exactly what you see on a billiards table.

Problem 4: unequal masses, elastic — the light body bounces back

A 1 kg ball moving at 6 m/s strikes a stationary 2 kg ball (e = 1). What happens?

The formulas give v₁′ = -2 m/s (negative, so the lighter ball rebounds backward) and v₂′ = 4 m/s. Checking kinetic energy: 18 J before, 18 J after — exactly equal, as an elastic collision requires.

Problem 5: a large kinetic-energy loss in a perfectly inelastic collision

A 4 kg cart moving at 3 m/s and a 2 kg cart moving the opposite way at -3 m/s (heading toward each other) collide and stick (e = 0). What is their common velocity, and how much kinetic energy is lost?

v_common = (m₁u₁+m₂u₂)/(m₁+m₂) = 1 m/s. Kinetic energy was 27 J before and only 3 J after sticking — 24 J is lost to heat, sound and deformation. Head-on collisions that stick lose the largest share of kinetic energy, and this is exactly that case.

Problem 6: finding the coefficient of restitution from measurements

An experiment records velocities before impact as u₁ = 6 m/s, u₂ = 0 m/s, and after impact as v₁′ = 1 m/s, v₂′ = 4 m/s. What is this collision's coefficient of restitution?

e = (v₂′ − v₁′)/(u₁ − u₂) = (4 − 1)/(6 − 0) = 0.50. This is exactly how e is measured for a real collision in a lab, without ever needing to know either mass.

Problem 7: a gun's recoil

A 5 kg gun fires a 0.05 kg bullet at 200 m/s. How fast does the gun recoil?

Total momentum was zero before firing, so it must stay zero afterward: 0 = m_gun v_gun + m_bullet v_bullet. Solving, v_gun = −(m_bullet × v_bullet)/m_gun = -2.0 m/s. The minus sign shows the gun moves opposite to the bullet — that recoil kick, an application of momentum conservation with no collision involved at all.

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.

e	v₁′	v₂′	Kinetic energy (after)	% of the original kinetic energy
0.00	1.00 m/s	1.00 m/s	2.50 J	14%
0.25	0.25 m/s	1.50 m/s	3.44 J	20%
0.50	-0.50 m/s	2.00 m/s	6.25 J	36%
0.75	-1.25 m/s	2.50 m/s	10.94 J	63%
1.00	-2.00 m/s	3.00 m/s	17.50 J	100%

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.

A typical exam-style question

Setup: A 2 kg cart moving at 4 m/s collides with a 3 kg cart moving the opposite way at 1 m/s. After the collision, the two carts are measured moving at v₁′ = -0.50 m/s and v₂′ = 2.00 m/s.

Questions typically asked: (a) State the law of conservation of momentum. (b) Define the coefficient of restitution. (c) Find this collision's coefficient of restitution. (d) Comment on whether this collision is elastic, inelastic, or perfectly inelastic.

For (c) and (d): substituting into e = (v₂′−v₁′)/(u₁−u₂) gives e = 0.50, which sits strictly between 0 and 1 — so the collision is inelastic but not perfectly inelastic, and some kinetic energy is lost, but the carts do not move together.

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

What does the law of conservation of momentum state?

When no net external force acts on a system, its total momentum stays exactly the same before and after any collision or interaction.

What is the difference between an elastic and an inelastic collision?

In an elastic collision (e = 1), both momentum and kinetic energy are conserved. In an inelastic collision (e < 1), momentum is still conserved, but some kinetic energy is lost to heat, sound or deformation.

What is the coefficient of restitution?

The coefficient of restitution, e, is the ratio of the relative speed at which two bodies separate after a collision to the relative speed at which they approached before it: e = (v₂′−v₁′)/(u₁−u₂).

What happens in a perfectly inelastic collision (e = 0)?

The two bodies do not separate after impact; they move together at one shared velocity, v₁′ = v₂′ = (m₁u₁+m₂u₂)/(m₁+m₂). This is where the largest possible fraction of kinetic energy is lost.

What are the final velocities in the simulation's default collision (m₁ = 2 kg, u₁ = 4 m/s, m₂ = 3 kg, u₂ = -1 m/s, e = 0.50)?

Substituting into the formulas gives v₁′ = -0.50 m/s and v₂′ = 2.00 m/s. Momentum is 5 kg·m/s both before and after, while kinetic energy falls from 17.50 J to 6.25 J.

What happens when two equal masses collide elastically?

If one is initially at rest, the moving one stops completely and the one that was at rest moves off with exactly the first one's original velocity — the velocities simply swap. This is what happens on a billiards table.

Can the coefficient of restitution be found from velocities alone, without knowing the masses?

Yes. Measuring the velocities before and after a collision and substituting into e = (v₂′−v₁′)/(u₁−u₂) gives e directly; the masses never enter that formula.

Where does the lost kinetic energy go in an inelastic collision?

It is not destroyed — just like momentum, energy is conserved overall. The kinetic energy "lost" from the bodies' motion is converted into heat, sound and permanent deformation of the colliding objects.

Is a gun's recoil an example of conservation of momentum?

Yes. The total momentum was zero before firing, so it must still be zero afterward — if the bullet moves forward, the gun is forced to move backward by exactly the amount needed to keep the total at zero.