Momentum and Collisions
3. Conservation of Momentum
Learning outcomes
- I can state the law of conservation of momentum.
- I can identify isolated systems.
- I can apply momentum conservation to simple situations.
- I can calculate unknown momenta after interactions.
- I can explain momentum transfer between objects.
What Is Conservation of Momentum?
Momentum describes the motion of an object and depends on both its mass and velocity.
For a single object:
p = mv
where:
- p = momentum in kg·m/s
- m = mass in kg
- v = velocity in m/s
Momentum is a vector quantity, so direction matters.
During interactions such as collisions and explosions, momentum can move from one object to another. However, under the correct conditions, the total momentum of the system remains constant.
This principle is called the law of conservation of momentum.
The Law of Conservation of Momentum
The law of conservation of momentum states:
The total momentum of an isolated system remains constant.
In other words:
total momentum before an interaction = total momentum after the interaction
We can write this as:
Σp(before) = Σp(after)
The symbol Σ means "the sum of."
So we add the momentum of every object in the system.
What Is a System?
A system is the object or group of objects that we choose to study.
For example, imagine two carts colliding on a track.
We could define our system as:
cart A + cart B
During the collision, the carts exert forces on each other.
These are internal forces because they occur between objects inside our chosen system.
What Is an Isolated System?
An isolated system is a system in which there is no significant net external force or external impulse acting during the interaction being analyzed.
Examples that can often be approximated as isolated during a short interaction include:
- two carts colliding on a low-friction track
- two ice skaters pushing apart
- billiard balls colliding
- objects separating after an explosion
- two spacecraft interacting far from significant external influences
Real systems are rarely perfectly isolated, but external effects can sometimes be small enough to ignore during a short interaction.
Internal and External Forces
It is important to distinguish between internal and external forces.
Internal forces act between objects within the system.
For two colliding carts:
- cart A pushes cart B
- cart B pushes cart A
These forces transfer momentum between the carts.
External forces are exerted by objects outside the system.
Examples might include:
- friction from the floor
- air resistance
- a person pushing one of the carts
- an external motor
- gravity, when its impulse in the direction being studied is significant
Momentum conservation applies directly when the net external impulse is zero or negligible.
Why Internal Forces Do Not Change Total Momentum
During a collision, object A exerts a force on object B.
At the same time, object B exerts an equal and opposite force on object A.
This is Newton's Third Law.
Because the forces are equal and opposite and act over the same interaction time, the impulses are equal and opposite.
One object gains momentum while the other loses an equal amount.
Therefore:
momentum can be transferred within the system without changing the total momentum of the system.
Momentum Before and After an Interaction
For two objects:
Before:
total momentum = p₁ + p₂
After:
total momentum = p₁' + p₂'
Therefore:
p₁ + p₂ = p₁' + p₂'
Using p = mv:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
The primes simply indicate the velocities after the interaction.
Direction Matters
Because momentum is a vector, we must choose a positive direction.
For example:
right = positive
left = negative
An object moving:
5 m/s right → +5 m/s
5 m/s left → −5 m/s
The signs must be included when calculating total momentum.
Example 1: One Moving Cart Hits a Stationary Cart
A 2 kg cart moves right at 4 m/s.
A 2 kg cart is stationary.
Before the collision:
Cart A:
p = mv
p = 2 × 4 = 8 kg·m/s
Cart B:
p = 2 × 0 = 0 kg·m/s
Total:
8 kg·m/s
Suppose after the collision cart A stops and cart B moves right.
Conservation of momentum requires:
total momentum after = 8 kg·m/s
Therefore:
2v = 8
v = 4 m/s
Cart B moves right at:
4 m/s
In this idealized example, momentum has effectively transferred from cart A to cart B.
Momentum Transfer
It is useful to think about momentum as being transferred between interacting objects.
Suppose:
Object A initially has:
+20 kg·m/s
Object B initially has:
0 kg·m/s
After the interaction:
Object A has:
+8 kg·m/s
Because the system's total momentum remains +20 kg·m/s, object B must have:
+12 kg·m/s
Object A lost:
12 kg·m/s
Object B gained:
12 kg·m/s
Total momentum remains unchanged.
Momentum Is Not Used Up
Momentum is not "used up" during a collision.
Instead, it can be:
- transferred between objects
- redistributed among several objects
- divided between different directions
As long as the system is isolated:
total momentum remains constant.
Example 2: Two Objects Moving in the Same Direction
A 3 kg cart moves right at 5 m/s.
A 2 kg cart moves right at 2 m/s.
Initial momentum:
Cart 1:
p₁ = 3 × 5 = 15 kg·m/s
Cart 2:
p₂ = 2 × 2 = 4 kg·m/s
Total:
p(total) = 15 + 4 = 19 kg·m/s
Suppose they collide and stick together.
Combined mass:
3 + 2 = 5 kg
Therefore:
5v = 19
v = 3.8 m/s
The combined carts move right at:
3.8 m/s
Objects That Stick Together
When two objects collide and stick together, they have the same final velocity.
The momentum equation becomes:
m₁v₁ + m₂v₂ = (m₁ + m₂)v
This type of collision is called a perfectly inelastic collision.
Momentum is conserved in an isolated system, but kinetic energy is not necessarily conserved.
Momentum and Kinetic Energy Are Different
This distinction is extremely important.
In an isolated collision:
total momentum is conserved.
But:
total kinetic energy may or may not be conserved.
Some kinetic energy can be transformed into:
- thermal energy
- sound
- deformation
- vibration
Therefore:
momentum conservation does not mean kinetic energy conservation.
Elastic Collisions
An elastic collision is one in which both:
- total momentum is conserved
- total kinetic energy is conserved
Ideal elastic collisions are useful models in physics.
Collisions between certain particles and approximately elastic collisions between objects such as billiard balls can illustrate this behaviour.
Inelastic Collisions
In an inelastic collision:
- momentum is conserved for an isolated system
- kinetic energy is not conserved as kinetic energy
Some kinetic energy is transformed into other forms.
If the objects stick together, the collision is perfectly inelastic.
Example 3: Opposite Directions
A 4 kg cart moves right at 3 m/s.
A 2 kg cart moves left at 5 m/s.
Choose:
right = positive
Therefore:
v₁ = +3 m/s
v₂ = −5 m/s
Momentum of cart 1:
p₁ = 4(+3) = +12 kg·m/s
Momentum of cart 2:
p₂ = 2(−5) = −10 kg·m/s
Total:
p(total) = +12 − 10
p(total) = +2 kg·m/s
The system therefore has a small net momentum to the right.
If the Carts Stick Together
The total mass becomes:
4 + 2 = 6 kg
Conservation of momentum:
6v = +2
Therefore:
v = +0.33 m/s
The positive answer means the combined carts move:
to the right
at approximately:
0.33 m/s
Why Signs Are Essential
If we ignored direction and simply added:
12 + 10 = 22 kg·m/s
we would obtain the wrong answer.
Momentum has both:
magnitude and direction
so vector signs must be included.
Example 4: Finding an Unknown Momentum
Two objects have a total initial momentum of:
+30 kg·m/s
After the collision, object A has:
+18 kg·m/s
What is object B's momentum?
Conservation of momentum:
30 = 18 + pB
Therefore:
pB = 12 kg·m/s
The positive sign indicates that object B's momentum is in the positive direction.
Example 5: Finding an Unknown Velocity
A 4 kg cart moves right at 6 m/s and collides with a stationary 2 kg cart.
After the collision, the 4 kg cart moves right at 3 m/s.
Find the velocity of the 2 kg cart.
Initial momentum:
p(initial) = (4)(6) + (2)(0)
p(initial) = 24 kg·m/s
Final momentum:
p(final) = (4)(3) + (2)v
Therefore:
24 = 12 + 2v
12 = 2v
v = 6 m/s
The second cart moves right at:
6 m/s
A Momentum Table Can Help
For complicated problems, organize the information before calculating.
For example:
Object A:
Mass = 4 kg
Initial velocity = +6 m/s
Final velocity = +3 m/s
Object B:
Mass = 2 kg
Initial velocity = 0 m/s
Final velocity = unknown
Then calculate the momentum of each object and apply:
Σp(before) = Σp(after)
This reduces sign and substitution errors.
Example 6: Collision and Rebound
A 1 kg ball moves right at 8 m/s.
After hitting another object, it rebounds left at 3 m/s.
Initial momentum:
pᵢ = 1(+8) = +8 kg·m/s
Final momentum:
p_f = 1(−3) = −3 kg·m/s
Change in momentum:
Δp = p_f − pᵢ
Δp = −3 − 8
Δp = −11 kg·m/s
The ball has experienced a large momentum change because its direction changed.
The other object or the wider environment receives an equal and opposite momentum change when the complete isolated system is considered.
Conservation of Momentum and Impulse
Impulse is related to momentum change:
J = Δp
During an interaction:
Object A experiences an impulse from object B.
Object B experiences an equal and opposite impulse from object A.
Therefore:
ΔpA = −ΔpB
This is another way of expressing momentum transfer.
Why Momentum Is Conserved
Newton's Third Law tells us that interacting objects exert equal and opposite forces.
If the interaction lasts for the same time Δt:
FAΔt = −FBΔt
Since impulse equals change in momentum:
ΔpA = −ΔpB
Therefore:
ΔpA + ΔpB = 0
The total momentum does not change.
This provides a connection between:
Newton's Third Law → impulse → conservation of momentum
Explosions and Separation
Momentum conservation also applies when objects move apart.
Imagine two carts initially connected by a compressed spring.
Initially both are stationary.
Total momentum:
0 kg·m/s
When released, the spring pushes the carts apart.
If the system is isolated:
total momentum after must also equal zero.
Therefore:
p₁ + p₂ = 0
or:
p₁ = −p₂
The carts have equal-magnitude but opposite momenta.
Example 7: Two Skaters Push Apart
Two ice skaters are initially stationary.
Skater A has mass:
60 kg
Skater B has mass:
40 kg
After pushing apart, the 60 kg skater moves left at 2 m/s.
Choose right as positive.
Therefore:
vA = −2 m/s
Initial total momentum:
0
After:
0 = (60)(−2) + (40)vB
0 = −120 + 40vB
40vB = 120
vB = +3 m/s
The 40 kg skater moves right at:
3 m/s
Notice that the lighter skater moves faster.
Equal Momentum Does Not Mean Equal Velocity
In the skater example:
60 kg × 2 m/s = 120 kg·m/s
40 kg × 3 m/s = 120 kg·m/s
Their momentum magnitudes are equal.
But their speeds are different.
Because:
p = mv
a smaller mass requires a larger speed to have the same momentum magnitude.
Example 8: Explosion from Rest
A 10 kg object initially at rest breaks into two pieces.
One piece has mass:
6 kg
and moves right at:
4 m/s
Its momentum is:
p = 6 × 4 = +24 kg·m/s
The original total momentum was zero.
Therefore, the other piece must have:
−24 kg·m/s
If its mass is 4 kg:
4v = −24
v = −6 m/s
Therefore, the second piece moves left at:
6 m/s
Recoil
Recoil is another application of momentum conservation.
When one part of a system is accelerated in one direction, another part can gain momentum in the opposite direction.
A useful non-weapon example is a balloon releasing air.
Air moves backward.
The balloon moves forward.
The momenta are in opposite directions.
Rockets and Momentum
Rockets also demonstrate momentum conservation.
A rocket expels exhaust gases backward at high velocity.
The gases gain backward momentum.
The rocket gains forward momentum.
This allows a rocket to accelerate even in space.
It does not need to push against the air.
Instead, momentum is exchanged between:
rocket + exhaust gases
Collisions Between Vehicles
Momentum conservation is useful when analyzing vehicle collisions.
If external impulses during the short collision are small compared with the collision forces, investigators can relate:
- vehicle masses
- directions
- velocities before collision
- velocities after collision
using conservation of momentum.
Real accident reconstruction is more complicated because investigators must also consider braking, friction, rotation, deformation, measurement uncertainty, and other evidence.
Billiards and Momentum
Billiard balls provide a familiar example.
When a moving ball strikes another ball, momentum can be transferred between them.
For a straight-line collision between equal masses, one ball may transfer a large fraction of its momentum to the other.
In two dimensions, momentum must be conserved separately in each direction:
Σpₓ(before) = Σpₓ(after)
Σpᵧ(before) = Σpᵧ(after)
Newton's Cradle
A Newton's cradle provides a striking demonstration of momentum transfer.
When one ball strikes the others, interactions transmit momentum through the system and a ball at the opposite end moves.
The real motion involves both momentum and energy transfer, and real cradles lose some mechanical energy through sound, deformation, and other processes.
When Is Momentum Not Conserved for the Chosen System?
Momentum conservation depends on how the system boundary is chosen.
Suppose a ball hits a wall.
If the system is only:
the ball
its momentum changes because the wall exerts an external force on it.
So the ball alone is not an isolated system.
But if we expand the system to include:
ball + wall + Earth
the momentum transferred to the wall and Earth becomes part of the system.
Total momentum is still conserved.
This demonstrates why defining the system is essential.
External Impulse
A more general relationship is:
Δp(system) = J_external
Therefore:
If:
J_external = 0
then:
Δp(system) = 0
and:
p_initial = p_final
This is the deeper condition behind momentum conservation.
Approximate Isolation
Many classroom situations are not perfectly isolated.
For example, a cart may experience some friction.
However, if:
- the collision happens very quickly
- friction is relatively small
- collision forces are much larger than external forces
then the external impulse during the collision may be negligible.
We can then treat the system as approximately isolated.
This is common in real physics.
A Reliable Problem-Solving Strategy
When solving conservation-of-momentum problems:
Step 1: Define the system.
Which objects are included?
Step 2: Decide whether momentum conservation is appropriate.
Is the net external impulse negligible?
Step 3: Choose a positive direction.
For example:
right = positive.
Step 4: Record masses and velocities.
Include negative signs for motion in the opposite direction.
Step 5: Calculate initial momentum.
Use:
p = mv
Step 6: Write the conservation equation.
Σp(before) = Σp(after)
Step 7: Substitute known values.
Step 8: Solve for the unknown.
Step 9: Interpret the sign.
Positive or negative tells you the direction.
Step 10: Check the answer.
Verify that total momentum before equals total momentum after.
Worked Example 9
A 5 kg cart moving right at 4 m/s collides with a stationary 3 kg cart.
Afterward, the 5 kg cart moves right at 1 m/s.
Find the second cart's velocity.
Initial momentum:
pᵢ = (5)(4) + (3)(0)
pᵢ = 20 kg·m/s
Final momentum:
p_f = (5)(1) + 3v
Conservation:
20 = 5 + 3v
15 = 3v
v = 5 m/s
Therefore, the second cart moves:
5 m/s to the right
Check:
Before:
20 kg·m/s
After:
5 + 15 = 20 kg·m/s
Momentum is conserved.
Worked Example 10
A 2 kg cart moving right at 6 m/s collides and sticks to a 4 kg cart moving left at 2 m/s.
Take right as positive.
Initial momentum:
pᵢ = (2)(+6) + (4)(−2)
pᵢ = 12 − 8
pᵢ = +4 kg·m/s
Combined mass:
6 kg
Therefore:
6v = 4
v = +0.67 m/s
The carts move together at approximately:
0.67 m/s to the right
Worked Example 11
Two objects initially at rest push apart.
Object A:
mass = 3 kg
velocity = +8 m/s
Object B:
mass = 6 kg
velocity = unknown
Initial momentum:
0
After:
0 = (3)(8) + 6v
0 = 24 + 6v
v = −4 m/s
Object B moves:
4 m/s in the opposite direction.
Worked Example 12: Unknown Mass
Two objects initially at rest separate.
A 2 kg object moves left at 6 m/s.
The second object moves right at 3 m/s.
Find its mass.
Take right as positive.
Initial momentum:
0
After:
0 = (2)(−6) + m(3)
0 = −12 + 3m
3m = 12
m = 4 kg
Momentum Conservation in Graphical Form
Momentum can also be represented with arrows.
The length of an arrow can represent the magnitude of momentum.
Before an interaction, the momentum vectors add to a total vector.
After the interaction, the individual vectors may change.
But for an isolated system:
the total vector must remain the same.
This becomes especially important in two-dimensional collisions.
Momentum Transfer and Newton's Third Law
Suppose object A loses:
6 kg·m/s
of momentum during a collision.
Then object B must gain:
6 kg·m/s
in the corresponding opposite-change sense for the two-object isolated system.
We can express this as:
ΔpA = −ΔpB
This is why momentum transfer between objects does not change the system's total momentum.
Did You Know?
Conservation of momentum applies far beyond collisions between everyday objects.
Particle physicists use momentum conservation when studying collisions between subatomic particles.
If the measured momentum after an interaction appears not to match the visible particles, scientists can investigate whether additional particles carried away momentum.
Conservation laws are therefore powerful tools for studying things that cannot always be observed directly.
Common Mistakes
Mistake 1: Conserving the momentum of each object separately
Individual objects can gain or lose momentum.
It is the total momentum of the isolated system that remains constant.
Mistake 2: Ignoring direction
Momentum is a vector.
Opposite directions must have opposite signs.
Mistake 3: Assuming stationary objects have no mass
A stationary object still has mass.
Its momentum is zero because:
v = 0
Mistake 4: Forgetting to add the masses when objects stick
If objects stick together:
final mass = m₁ + m₂
Mistake 5: Assuming kinetic energy is always conserved
Momentum is conserved in an isolated system.
Kinetic energy is conserved only in elastic collisions.
Mistake 6: Assuming momentum disappears in a collision
Momentum is transferred and redistributed.
It is not destroyed.
Mistake 7: Forgetting to define the system
Whether a force is internal or external depends on the chosen system boundary.
Mistake 8: Assuming every real system is perfectly isolated
Real systems may experience friction, air resistance, or other external forces.
Momentum conservation can still be a useful approximation when the external impulse is negligible.
Key Terms
Momentum: A vector quantity equal to mass multiplied by velocity.
Conservation of momentum: The principle that total momentum remains constant in an isolated system.
System: The object or collection of objects being studied.
Isolated system: A system experiencing no significant net external impulse during the interaction.
Internal force: A force between objects within the system.
External force: A force exerted on the system by something outside it.
External impulse: The impulse produced by external forces acting on a system.
Collision: An interaction in which objects exert forces on each other over a short time.
Elastic collision: A collision in which both momentum and kinetic energy are conserved.
Inelastic collision: A collision in which momentum is conserved but kinetic energy is not conserved as kinetic energy.
Perfectly inelastic collision: A collision in which objects stick together.
Recoil: Motion in one direction resulting from momentum being carried in the opposite direction.
Momentum transfer: The change in momentum of interacting objects caused by forces between them.
Key Equations
Momentum:
p = mv
Conservation of momentum:
Σp(before) = Σp(after)
For two objects:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
If two objects stick together:
m₁v₁ + m₂v₂ = (m₁ + m₂)v
For an explosion initially at rest:
0 = m₁v₁ + m₂v₂
Momentum transfer between two objects in an isolated system:
Δp₁ = −Δp₂
Impulse and system momentum:
J_external = Δp_system
Key Takeaways
- Momentum is calculated using p = mv.
- Momentum is a vector, so direction matters.
- The law of conservation of momentum states that the total momentum of an isolated system remains constant.
- Therefore, total momentum before an interaction equals total momentum after it.
- A system is the group of objects chosen for analysis.
- Momentum conservation requires the net external impulse on the system to be zero or negligible.
- Forces between objects within the system are internal forces.
- Internal forces can transfer momentum between objects without changing the system's total momentum.
- When one object loses momentum, another part of the isolated system gains an equal amount of momentum in the appropriate vector direction.
- Objects moving in opposite directions must be assigned opposite velocity signs.
- Objects that stick together have the same final velocity.
- Momentum is conserved in both elastic and inelastic collisions when the system is isolated.
- Kinetic energy is conserved only in elastic collisions.
- Objects initially at rest can move apart while maintaining a total momentum of zero.
- Lighter objects often move faster than heavier objects when they separate with equal and opposite momenta.
- Recoil, rockets, collisions, billiards, skaters, and particle interactions can all be analyzed using conservation of momentum.
- Real systems can often be treated as approximately isolated when external impulses are very small during the interaction.
- A reliable momentum-conservation solution follows:
define the system → choose a positive direction → calculate momentum before → calculate momentum after → set totals equal → solve → interpret direction → check conservation.