Last Updated: September 5, 2026
Quick Summary: Conservation of Momentum
The law of conservation of momentum states that the total linear momentum of an isolated system remains constant. In other words, total momentum before an interaction equals total momentum after the interaction.
Key formulas
Linear momentum:
[
p = mv
]
Conservation of momentum:
[
\sum p_i = \sum p_f
]
For two objects:
[
m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2
]
For a perfectly inelastic collision where two objects stick together:
[
m_1u_1 + m_2u_2 = (m_1+m_2)v
]
Impulse-momentum theorem:
[
J = \Delta p = F\Delta t
]
Momentum is conserved when the net external impulse on the chosen system is zero. This applies to elastic collisions, inelastic collisions, explosions, recoil, and many other interactions.
Introduction
The conservation of momentum is one of the most useful principles in classical mechanics. It allows you to determine unknown velocities and momentum changes without calculating every force acting during an interaction.
The central idea is simple:
[
\boxed{\text{Total momentum before}=\text{Total momentum after}}
]
This principle is especially useful for collisions, explosions, recoil problems, two-dimensional interactions, and impulse calculations.
Because momentum is a vector quantity, direction matters. A velocity moving to the right can be assigned a positive sign, while a velocity moving to the left is normally assigned a negative sign.
In this guide, you will learn the law of conservation of momentum, conservation of momentum formulas, collision equations, impulse, 2D momentum, worked examples, and common exam mistakes.
What Is the Law of Conservation of Momentum?
The law of conservation of momentum states:
If the net external impulse on a system is zero, the total linear momentum of that system remains constant.
Mathematically:
[
\boxed{\sum p_i = \sum p_f}
]
For two objects:
[
\boxed{m_1u_1+m_2u_2=m_1v_1+m_2v_2}
]
Where:
- (m_1,m_2) = masses
- (u_1,u_2) = initial velocities
- (v_1,v_2) = final velocities
- (p_i) = total initial momentum
- (p_f) = total final momentum
When Is Momentum Conserved?
Momentum is conserved for a system when the net external impulse is zero during the interaction.
This commonly occurs when:
- external forces are negligible during a short collision
- the system is isolated
- internal forces dominate the interaction
- all relevant interacting objects are included in the system
For example, two balls can exert large forces on each other during a collision, but those forces are internal to the two-ball system. Their momentum changes are equal and opposite, so the total momentum remains constant.
Linear Momentum Formula
Before using conservation of momentum, you need to understand linear momentum.
The formula is:
[
\boxed{p=mv}
]
Where:
- (p) = momentum
- (m) = mass
- (v) = velocity
The SI unit of momentum is:
[
\boxed{\text{kg}\cdot\text{m/s}}
]
Momentum can also be expressed in:
[
\boxed{\text{N}\cdot\text{s}}
]
because:
[
1\text{ N}\cdot\text{s}=1\text{ kg}\cdot\text{m/s}
]
Example
A (2,\text{kg}) ball travels east at (5,\text{m/s}).
Taking east as positive:
[
p=mv
]
[
p=(2)(5)=10,\text{kg}\cdot\text{m/s}
]
So the ball has:
[
\boxed{p=+10,\text{kg}\cdot\text{m/s}}
]
If the same ball traveled west at (5,\text{m/s}), its momentum would be:
[
\boxed{-10,\text{kg}\cdot\text{m/s}}
]
The negative sign represents direction.
Conservation of Momentum Formula
The most important conservation of momentum equation is:
[
\boxed{\sum p_i=\sum p_f}
]
For a two-object collision:
[
\boxed{m_1u_1+m_2u_2=m_1v_1+m_2v_2}
]
This is the main conservation of momentum formula used in one-dimensional collision problems.
Momentum Before and After a Collision
The problem-solving process is:
- Choose a positive direction.
- Identify the masses.
- Write the initial velocities.
- Assign positive and negative signs.
- Calculate total initial momentum.
- Write the final momentum.
- Set initial momentum equal to final momentum.
- Solve for the unknown quantity.
- Check the direction and units.
Why Is Momentum Conserved?
Momentum conservation can be understood through Newton’s Third Law.
When two objects collide, each object exerts a force on the other. These forces are equal in magnitude and opposite in direction.
Because the interaction forces act over the same time interval, the impulses are also equal and opposite:
[
\Delta p_1=-\Delta p_2
]
Therefore:
[
\Delta p_1+\Delta p_2=0
]
So the total momentum change of the two-object system is zero.
This is why the momentum lost by one object is gained by the other, provided the net external impulse on the system is negligible.
Types of Collisions
There are three important collision classifications.
| Collision | Momentum | Kinetic Energy | What Happens? |
|---|---|---|---|
| Elastic | Conserved | Conserved | Objects separate without net kinetic-energy loss |
| Inelastic | Conserved | Not conserved | Some kinetic energy becomes heat, sound, or deformation |
| Perfectly inelastic | Conserved | Not conserved | Objects stick together after collision |
Is Momentum Conserved in Elastic Collisions?
Yes.
For an isolated system, momentum is conserved in an elastic collision.
Kinetic energy is also conserved:
[
KE_i=KE_f
]
Therefore, an elastic collision has both:
[
\boxed{p_i=p_f}
]
and
[
\boxed{KE_i=KE_f}
]
Is Momentum Conserved in Inelastic Collisions?
Yes.
Momentum is still conserved in an isolated system during an inelastic collision.
However, kinetic energy is not conserved.
Some kinetic energy can be transformed into:
- heat
- sound
- deformation
- internal energy
Therefore:
[
\boxed{p_i=p_f}
]
but generally:
[
\boxed{KE_i\neq KE_f}
]
Perfectly Inelastic Collision Formula
In a perfectly inelastic collision, the objects stick together and share one final velocity.
The equation becomes:
[
\boxed{m_1u_1+m_2u_2=(m_1+m_2)v}
]
Solving for the final velocity:
[
\boxed{v=\frac{m_1u_1+m_2u_2}{m_1+m_2}}
]
Conservation of Momentum in Collisions
Consider a (1000,\text{kg}) car traveling east at (20,\text{m/s}) that collides with a stationary (1500,\text{kg}) truck. The vehicles stick together.
Take east as positive.
Initial momentum:
[
p_i=(1000)(20)+(1500)(0)
]
[
p_i=20,000,\text{kg}\cdot\text{m/s}
]
Because the vehicles stick together:
[
20,000=(1000+1500)v
]
[
20,000=2500v
]
[
\boxed{v=8,\text{m/s east}}
]
The final velocity is therefore (8,\text{m/s}) east.
Elastic Collision Formula
For a one-dimensional elastic collision, both momentum and kinetic energy are conserved.
The two equations are:
[
m_1u_1+m_2u_2=m_1v_1+m_2v_2
]
and:
\frac12m_1v_1^2+\frac12m_2v_2^2
]
For a special case where object 2 is initially stationary:
[
u_2=0
]
the final velocities can be written as:
[
\boxed{v_1=\frac{m_1-m_2}{m_1+m_2}u_1}
]
[
\boxed{v_2=\frac{2m_1}{m_1+m_2}u_1}
]
These equations apply to a one-dimensional elastic collision with object 2 initially at rest.
Equal-Mass Elastic Collision
A useful special case occurs when two objects have equal masses.
Suppose:
- (m_1=m_2)
- object 2 is initially at rest
- the collision is elastic
The moving object transfers its velocity to the stationary object.
For example:
[
u_1=6,\text{m/s},\qquad u_2=0
]
After the collision:
[
\boxed{v_1=0}
]
[
\boxed{v_2=6,\text{m/s}}
]
Momentum is conserved:
[
p_i=2(6)+2(0)=12
]
[
p_f=2(0)+2(6)=12
]
Impulse-Momentum Theorem
Impulse describes how a force changes momentum.
The impulse formula is:
[
\boxed{J=F\Delta t}
]
The impulse-momentum theorem gives:
[
\boxed{J=\Delta p}
]
Therefore:
[
\boxed{F\Delta t=\Delta p}
]
For constant mass:
[
\boxed{F\Delta t=m(v_f-v_i)}
]
Impulse is measured in:
[
\boxed{\text{N}\cdot\text{s}}
]
which is equivalent to:
[
\boxed{\text{kg}\cdot\text{m/s}}
]
The Physics Classroom also describes impulse as force acting over time to produce a change in momentum.
Example: Impulse
A (0.5,\text{kg}) ball initially at rest receives an impulse of (12,\text{N}\cdot\text{s}).
Using:
[
J=m\Delta v
]
[
12=(0.5)\Delta v
]
[
\boxed{\Delta v=24,\text{m/s}}
]
Because the ball started from rest:
[
\boxed{v_f=24,\text{m/s}}
]
Impulse, Force, and Collision Time
From:
[
F=\frac{\Delta p}{\Delta t}
]
for a given momentum change, increasing the collision time reduces the average force.
This principle is used in:
- airbags
- vehicle crumple zones
- helmets
- protective padding
- sports equipment
For example, an airbag increases the time over which a passenger’s momentum changes during a crash, reducing the average force on the passenger.
2D Conservation of Momentum
Momentum is a vector, so two-dimensional collision problems must be handled separately along perpendicular axes.
For the x-direction:
[
\boxed{\sum p_{x,i}=\sum p_{x,f}}
]
For the y-direction:
[
\boxed{\sum p_{y,i}=\sum p_{y,f}}
]
For two objects:
m_1v_{1x}+m_2v_{2x}
]
and:
m_1v_{1y}+m_2v_{2y}
]
The key rule is:
[
\boxed{\text{Conserve momentum independently along x and y}}
]
2D Example
A (2,\text{kg}) ball travels east at (5,\text{m/s}) and collides with a stationary (3,\text{kg}) ball. After the collision, the (2,\text{kg}) ball moves north at (3,\text{m/s}).
For the x-direction:
[
2(5)=3v_x
]
[
v_x=\frac{10}{3}=3.33,\text{m/s east}
]
For the y-direction:
[
0=2(3)+3v_y
]
[
v_y=-2,\text{m/s}
]
So the second ball has velocity components:
[
\boxed{v_x=3.33,\text{m/s east}}
]
[
\boxed{v_y=2,\text{m/s south}}
]
Its speed is approximately:
[
v=\sqrt{3.33^2+2^2}
]
[
\boxed{v\approx3.89,\text{m/s}}
]
Conservation of Momentum in Explosions and Recoil
Momentum conservation also applies when an object separates into multiple pieces.
If the system is initially at rest:
[
p_i=0
]
Therefore:
[
\boxed{\sum p_f=0}
]
Recoil Example
A (4,\text{kg}) rifle fires a (0.02,\text{kg}) projectile at (400,\text{m/s}).
Initially:
[
p_i=0
]
After firing:
[
0=(0.02)(400)+(4)v_r
]
[
0=8+4v_r
]
[
4v_r=-8
]
[
\boxed{v_r=-2,\text{m/s}}
]
The negative sign means the rifle moves backward.
Rocket Propulsion and Momentum
A rocket can accelerate even in space because it ejects exhaust gases in the opposite direction.
The rocket and expelled gas can be treated as parts of a system whose momentum is conserved when external forces are negligible.
This is why rocket propulsion does not require the rocket to push against air.
When Is Momentum Not Conserved for an Object?
Momentum is not necessarily conserved for a selected object if there is a significant external impulse.
For example, friction can change the momentum of a sliding object.
However, whether momentum is conserved depends on the system boundary.
If you consider only the sliding object, friction from the ground is external.
If you include the relevant interacting bodies and environment in a larger system, the total momentum accounting can change.
The correct question is therefore:
[
\boxed{\text{Is the net external impulse on the chosen system zero?}}
]
rather than simply asking whether friction exists.
Conservation of Momentum Worked Examples
Example 1: Basic Momentum
A (1,500,\text{kg}) car travels at (20,\text{m/s}).
[
p=mv
]
[
p=(1500)(20)
]
[
\boxed{p=30,000,\text{kg}\cdot\text{m/s}}
]
Example 2: Perfectly Inelastic Collision
A (3,\text{kg}) ball travels at (4,\text{m/s}) and hits a stationary (2,\text{kg}) ball. They stick together.
[
3(4)+2(0)=(3+2)v
]
[
12=5v
]
[
\boxed{v=2.4,\text{m/s}}
]
Example 3: Head-On Collision
A (4,\text{kg}) ball travels right at (6,\text{m/s}). A (2,\text{kg}) ball travels left at (3,\text{m/s}). They stick together.
Take right as positive:
[
4(6)+2(-3)=(4+2)v
]
[
24-6=6v
]
[
18=6v
]
[
\boxed{v=3,\text{m/s right}}
]
Example 4: Explosion
A stationary (5,\text{kg}) object explodes into a (3,\text{kg}) fragment moving right at (4,\text{m/s}) and a (2,\text{kg}) fragment.
Initial momentum is zero:
[
0=3(4)+2v
]
[
0=12+2v
]
[
\boxed{v=-6,\text{m/s}}
]
The (2,\text{kg}) fragment moves left at (6,\text{m/s}).
Example 5: Impulse
A (50,\text{N}) force acts on an object for (3,\text{s}).
[
J=F\Delta t
]
[
J=(50)(3)
]
[
\boxed{J=150,\text{N}\cdot\text{s}}
]
Therefore:
[
\boxed{\Delta p=150,\text{kg}\cdot\text{m/s}}
]
Example 6: Ball Bouncing From a Wall
A (0.4,\text{kg}) ball travels toward a wall at (8,\text{m/s}) and bounces back at (6,\text{m/s}).
Take motion toward the wall as positive:
[
v_i=8,\text{m/s}
]
[
v_f=-6,\text{m/s}
]
Therefore:
[
\Delta p=m(v_f-v_i)
]
[
\Delta p=0.4(-6-8)
]
[
\boxed{\Delta p=-5.6,\text{kg}\cdot\text{m/s}}
]
The negative sign represents the reversal in direction.
How to Solve Conservation of Momentum Problems
Use this checklist for exams and numerical problems.
Step 1: Define the system
Decide which objects are included in the system.
Step 2: Choose a positive direction
For one-dimensional problems, choose right or east as positive.
Step 3: Write the initial momentum
Use:
[
p=mv
]
for each object.
Step 4: Write the final momentum
Use the appropriate final velocities.
Step 5: Apply conservation
[
\sum p_i=\sum p_f
]
Step 6: Solve for the unknown
Rearrange the equation carefully.
Step 7: Check units and direction
Momentum should be in:
[
\text{kg}\cdot\text{m/s}
]
Velocity should be in:
[
\text{m/s}
]
A negative velocity indicates motion opposite to your chosen positive direction.
Common Conservation of Momentum Mistakes
1. Forgetting that momentum has direction
Momentum is a vector.
Do not treat opposite directions as positive quantities.
2. Assuming kinetic energy is always conserved
Kinetic energy is conserved in elastic collisions, but not generally in inelastic collisions.
3. Forgetting the second object
For a collision:
[
p_{\text{total}}=p_1+p_2
]
You must include every relevant object in the system.
4. Using speed instead of velocity
Velocity includes direction.
A velocity of (-5,\text{m/s}) is not the same as (+5,\text{m/s}).
5. Forgetting that objects stick together
For a perfectly inelastic collision:
[
v_1=v_2=v
]
so:
[
m_1u_1+m_2u_2=(m_1+m_2)v
]
6. Conserving momentum for the wrong system
If an external impulse acts on your selected system, total momentum of that system may change.
Always identify the system before writing the conservation equation.
Conservation of Momentum Formula Sheet
Linear momentum
[
\boxed{p=mv}
]
Conservation of momentum
[
\boxed{\sum p_i=\sum p_f}
]
Two-object collision
[
\boxed{m_1u_1+m_2u_2=m_1v_1+m_2v_2}
]
Perfectly inelastic collision
[
\boxed{m_1u_1+m_2u_2=(m_1+m_2)v}
]
Impulse
[
\boxed{J=F\Delta t}
]
Impulse-momentum theorem
[
\boxed{J=\Delta p}
]
Momentum change
[
\boxed{\Delta p=m(v_f-v_i)}
]
Elastic collision
[
\boxed{p_i=p_f}
]
and:
[
\boxed{KE_i=KE_f}
]
2D momentum
[
\boxed{\sum p_x{}_i=\sum p_x{}_f}
]
[
\boxed{\sum p_y{}_i=\sum p_y{}_f}
]
Frequently Asked Questions
What is the law of conservation of momentum?
The law of conservation of momentum states that the total linear momentum of an isolated system remains constant. Therefore:
[
\boxed{\sum p_i=\sum p_f}
]
What is the conservation of momentum formula?
For two interacting objects, the standard formula is:
[
\boxed{m_1u_1+m_2u_2=m_1v_1+m_2v_2}
]
When is momentum conserved?
Momentum is conserved when the net external impulse on the chosen system is zero. This commonly applies to isolated systems and many short-duration collision or explosion problems.
Is momentum conserved in all collisions?
For an isolated system, yes. Momentum is conserved in elastic, inelastic, and perfectly inelastic collisions.
Kinetic energy, however, is not conserved in general.
Is momentum conserved in an inelastic collision?
Yes. Total momentum is conserved in an isolated inelastic collision, although kinetic energy is not generally conserved.
Is momentum conserved in a perfectly inelastic collision?
Yes. Momentum is conserved if the system has zero net external impulse. The objects stick together, but their total momentum remains the same before and after the collision.
Is momentum conserved in an elastic collision?
Yes. In an isolated elastic collision, both total momentum and total kinetic energy are conserved.
How is momentum conserved?
During an interaction, internal forces produce equal and opposite momentum changes. If there is no net external impulse, these changes cancel when the entire system is considered.
What is the difference between momentum and impulse?
Momentum is:
[
p=mv
]
Impulse is the change in momentum:
[
J=\Delta p
]
For a constant force:
[
J=F\Delta t
]
What happens to momentum during an explosion?
If the system is isolated, total momentum remains constant. An initially stationary object has zero total momentum, so the fragments’ momenta must add to zero after the explosion.
Does friction always destroy momentum conservation?
Not necessarily. Friction can provide an external impulse to a selected system, but conservation depends on the complete system boundary and whether the net external impulse is zero.
Quick Exam Checklist
Before submitting a conservation of momentum answer, check:
- Did I define the system?
- Did I choose a positive direction?
- Did I assign signs to velocities?
- Did I calculate momentum using (p=mv)?
- Did I include every relevant object?
- Did I use (\sum p_i=\sum p_f)?
- Did I distinguish elastic from inelastic collisions?
- Did I check whether the objects stick together?
- Did I use separate x and y equations for a 2D problem?
- Did I include correct units?
- Does the final direction make physical sense?
Final Takeaway
The central idea behind conservation of momentum is:
[
\boxed{\text{Total momentum before}=\text{Total momentum after}}
]
when the net external impulse on the chosen system is zero.
The most important equations to remember are:
[
\boxed{p=mv}
]
[
\boxed{\sum p_i=\sum p_f}
]
[
\boxed{m_1u_1+m_2u_2=m_1v_1+m_2v_2}
]
[
\boxed{J=\Delta p=F\Delta t}
]
For perfectly inelastic collisions:
[
\boxed{m_1u_1+m_2u_2=(m_1+m_2)v}
]
Once you understand direction, system boundaries, and the difference between momentum and kinetic energy, conservation of momentum becomes one of the most reliable tools for solving collision, recoil, explosion, and impulse problems.