Momentum and Collisions
5. Explosions and Recoil
Learning outcomes
- I can explain recoil using momentum conservation.
- I can analyze explosions as momentum interactions.
- I can calculate velocities after explosions.
- I can explain propulsion systems using momentum principles.
- I can apply momentum conservation to real-world examples.
What Are Explosions and Recoil?
In physics, an explosion does not necessarily mean fire or a destructive event.
An explosion is any interaction in which parts of a system that were initially together move apart because stored energy is released.
Examples include:
- two carts pushed apart by a compressed spring
- two people on skateboards pushing away from each other
- a balloon releasing air
- a rocket ejecting exhaust gases
- fragments separating after an object breaks apart
The objects exert forces on each other and move in different directions.
If external forces are negligible during the interaction, the total momentum of the system is conserved.
Review: Momentum
Momentum depends on mass and velocity.
Momentum is measured in:
kg·m/s
Because velocity has direction, momentum is a vector quantity.
This means that direction is extremely important when analyzing explosions and recoil.
Conservation of Momentum
For an isolated system:
total momentum before = total momentum after
or:
Σp(before) = Σp(after)
For two objects:
m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f
This equation works for:
- collisions
- explosions
- recoil
- objects pushing apart
- many propulsion situations
The same conservation law applies even though the objects may behave very differently.
Explosions from Rest
Many introductory explosion problems begin with an object or system that is stationary.
If the system is initially at rest:
initial momentum = 0
Therefore:
final total momentum must also equal 0
For two objects moving apart:
m₁v₁ + m₂v₂ = 0
Therefore:
m₁v₁ = −m₂v₂
The negative sign tells us that the two momenta point in opposite directions.
Equal and Opposite Momentum
Suppose an object initially has zero momentum.
It separates into two pieces.
If one piece gains:
+20 kg·m/s
of momentum, the other must have:
−20 kg·m/s
of momentum.
Therefore:
total momentum = +20 + (−20) = 0
The two objects do not necessarily have equal velocities.
They have equal momentum magnitudes in this simple two-object case.
Why the Velocities Can Be Different
Remember:
p = mv
If two objects have equal momentum magnitudes but different masses, their velocities must be different.
For example:
Object A:
mass = 2 kg
momentum = +12 kg·m/s
v = 12 ÷ 2 = +6 m/s
Object B:
mass = 6 kg
momentum = −12 kg·m/s
v = −12 ÷ 6 = −2 m/s
The lighter object moves faster.
This gives us an important relationship:
smaller mass → larger speed
when the momentum magnitudes are equal.
Example 1: Two Carts Push Apart
Two carts are initially stationary.
A compressed spring between them is released.
Cart A:
mass = 3 kg
velocity = +4 m/s
Cart B:
mass = 6 kg
velocity = unknown
Initial momentum:
0 kg·m/s
After release:
0 = (3)(+4) + (6)v
0 = 12 + 6v
6v = −12
v = −2 m/s
Therefore, Cart B moves:
2 m/s in the opposite direction.
What Caused the Carts to Move?
Before release, energy was stored in the compressed spring as elastic potential energy.
When the spring was released:
elastic potential energy → kinetic energy
The spring exerted forces on both carts.
The carts gained opposite momenta.
The system's total momentum remained zero, but its total kinetic energy increased.
This is possible because momentum and energy are different conserved quantities.
Where Does the Kinetic Energy Come From?
An explosion can increase the kinetic energy of the objects.
This does not violate conservation of energy.
The kinetic energy comes from another form of stored energy.
Possible sources include:
- elastic potential energy
- chemical energy
- electrical energy
- pressure energy
- nuclear energy
For a compressed spring:
elastic potential energy → kinetic energy
For a rocket:
chemical energy → thermal energy + kinetic energy + other forms
Momentum vs Kinetic Energy
Suppose two carts initially sit at rest.
Initial momentum:
0
Initial kinetic energy:
0
After a spring pushes them apart:
Total momentum:
still 0
But total kinetic energy:
greater than 0
Where did the energy come from?
The spring's stored potential energy.
This illustrates why:
conservation of momentum does not mean kinetic energy must remain constant.
What Is Recoil?
Recoil is the motion of one part of a system in response to another part gaining momentum in the opposite direction.
The basic idea is:
one part moves one way → another part gains momentum the other way
If the system initially has zero momentum:
p₁ = −p₂
Recoil is therefore a direct consequence of momentum conservation.
Newton's Third Law and Recoil
Recoil can also be explained using Newton's Third Law.
During the interaction:
Object A exerts a force on Object B.
Object B exerts an equal and opposite force on Object A.
These forces act for the same time interval.
Therefore, the impulses are equal and opposite:
J₁ = −J₂
Since:
J = Δp
the momentum changes are also equal and opposite:
Δp₁ = −Δp₂
So Newton's Third Law and conservation of momentum describe the same interaction from complementary perspectives.
Example 2: Two Skaters Push Apart
Two skaters are initially stationary.
Skater A:
mass = 50 kg
Skater B:
mass = 75 kg
After pushing apart, Skater A moves left at 3 m/s.
Choose:
right = positive
Therefore:
vA = −3 m/s
Initial momentum:
0
Final momentum:
0 = (50)(−3) + (75)vB
0 = −150 + 75vB
75vB = 150
vB = +2 m/s
Skater B moves:
2 m/s to the right.
The lighter skater moves faster.
Example 3: Finding an Unknown Mass
Two carts initially at rest push apart.
Cart A:
mass = 4 kg
velocity = +6 m/s
Cart B:
velocity = −3 m/s
Find Cart B's mass.
Initial momentum:
0
Therefore:
0 = (4)(6) + m(−3)
0 = 24 − 3m
3m = 24
m = 8 kg
The slower cart has the greater mass.
Example 4: An Object Breaks into Two Pieces
A stationary 12 kg object separates into two pieces.
Piece A:
mass = 4 kg
velocity = +9 m/s
Piece B:
mass = 8 kg
velocity = unknown
Initial momentum:
0
Therefore:
0 = (4)(9) + (8)v
0 = 36 + 8v
8v = −36
v = −4.5 m/s
Piece B moves:
4.5 m/s in the opposite direction.
Checking the Momentum
Piece A:
pA = (4)(+9) = +36 kg·m/s
Piece B:
pB = (8)(−4.5) = −36 kg·m/s
Total:
+36 − 36 = 0
The original object was stationary, so the result satisfies conservation of momentum.
Explosions with Initial Motion
Not every explosion begins from rest.
Suppose an object is already moving when it separates.
Then:
initial momentum is not zero.
The correct relationship is still:
total momentum before = total momentum after
but we must calculate the original momentum first.
Example 5: Moving Object Separates
A 10 kg object moves right at 5 m/s.
It separates into two pieces.
Piece A:
mass = 4 kg
velocity = +8 m/s
Piece B:
mass = 6 kg
velocity = unknown
Initial momentum:
pᵢ = (10)(5)
pᵢ = +50 kg·m/s
After:
50 = (4)(8) + (6)v
50 = 32 + 6v
18 = 6v
v = +3 m/s
Piece B continues moving:
3 m/s to the right.
Notice that both pieces can move in the same direction.
Momentum conservation does not require explosion fragments to move in opposite directions.
Why Both Pieces Can Move Forward
Before the explosion, the original object already had forward momentum.
The explosion redistributes that momentum.
One piece might speed up while another slows down.
As long as:
total momentum after = total momentum before
momentum is conserved.
This is why it is essential to calculate the initial momentum rather than automatically setting it equal to zero.
More Than Two Fragments
An explosion can produce more than two moving objects.
The same rule applies:
Σp(before) = Σp(after)
For three fragments:
pᵢ = p₁ + p₂ + p₃
If the original object was stationary:
0 = p₁ + p₂ + p₃
Example 6: Three Fragments
A stationary object separates into three pieces along one line.
Piece A momentum:
+20 kg·m/s
Piece B momentum:
−8 kg·m/s
Find the momentum of Piece C.
Initial momentum:
0
Therefore:
0 = 20 − 8 + pC
0 = 12 + pC
pC = −12 kg·m/s
Piece C must carry:
12 kg·m/s to the left.
Momentum Vectors in Explosions
In more complex explosions, fragments may move in different directions.
Momentum must then be conserved separately in each dimension.
For two dimensions:
Σpₓ(before) = Σpₓ(after)
and:
Σpᵧ(before) = Σpᵧ(after)
For the current topic, however, most calculations can be handled as one-dimensional interactions.
Balloon Propulsion
A balloon provides a simple demonstration of recoil and propulsion.
Inflate a balloon and release it without tying the opening.
Air rushes backward out of the opening.
The balloon moves forward.
The system can be considered:
balloon + escaping air
The air gains momentum in one direction.
The balloon gains momentum in the opposite direction.
This is recoil.
Does the Balloon Push Against the Air?
No.
The balloon's motion is not fundamentally caused by pushing against the surrounding atmosphere.
The balloon pushes air out of itself.
The expelled air carries momentum backward.
The balloon gains forward momentum.
This becomes especially important when understanding rockets.
Rocket Propulsion
A rocket operates using the same fundamental principle.
Inside the rocket, chemical reactions produce hot, high-pressure gases.
The gases are expelled from the rocket at high velocity.
The exhaust gains momentum:
backward
The rocket gains momentum:
forward
Total momentum is conserved for the appropriately defined system.
Rockets Do Not Need Air
A common misconception is:
"Rockets push against the air."
They do not.
A rocket can operate in a vacuum.
The rocket interacts with its own expelled propellant.
The relevant system includes:
rocket + exhaust
Momentum carried backward by the exhaust is accompanied by forward momentum of the rocket.
This is why rockets work in space.
Momentum in Rocket Propulsion
A simplified picture is:
Before:
rocket + propellant moving together.
After some propellant is expelled:
exhaust → backward momentum
rocket → forward momentum
The momentum changes balance when the complete isolated system is considered.
In real rockets, the situation is more complicated because the rocket continuously loses mass as propellant is expelled.
This leads to more advanced equations of rocket motion.
Why Exhaust Speed Matters
Suppose a propulsion system expels a certain mass of gas.
Momentum is:
p = mv
Increasing the exhaust velocity increases the magnitude of momentum carried by the exhaust.
That produces a corresponding momentum change of the vehicle.
This is one reason high exhaust velocity is important in propulsion-system design.
Thrust and Momentum
Thrust is the force that accelerates a rocket or other propulsion system.
A simplified momentum idea is:
force = rate of change of momentum
or:
F = Δp/Δt
If exhaust carries momentum away rapidly, the rocket experiences a corresponding force.
For a steady idealized exhaust stream, thrust is closely related to:
mass flow rate × exhaust velocity
This provides the connection between momentum conservation and rocket thrust.
Jet Engines and Momentum
Jet engines also use momentum principles.
A jet engine takes in air and accelerates gases backward.
The backward-moving gases gain momentum.
The aircraft experiences forward thrust.
Unlike rockets, jet engines use oxygen from the atmosphere for combustion, so conventional jet engines cannot operate in the vacuum of space.
Propellers and Momentum
Propellers also create thrust by changing the momentum of a fluid.
An aircraft propeller accelerates air backward.
The aircraft gains forward momentum.
A boat propeller accelerates water backward.
The boat gains forward momentum.
The general pattern is:
fluid momentum backward → vehicle momentum forward
Water-Jet Propulsion
Some boats use water jets.
Water is drawn into the system and accelerated backward.
The expelled water carries backward momentum.
The boat gains forward momentum.
Again:
momentum transfer produces propulsion.
Squid and Octopus Propulsion
Momentum-based propulsion also appears in nature.
Squid and some other cephalopods can take water into their bodies and force it out through a narrow opening.
Water moves one way.
The animal accelerates the other way.
This is biological jet propulsion based on the same momentum principle.
Recoil in Everyday Situations
Recoil can be observed without any explosion.
Examples include:
- jumping from a stationary skateboard
- two people pushing apart on roller skates
- releasing an inflated balloon
- stepping from a small floating boat
- throwing an object while standing on low-friction wheels
In each case, momentum is transferred between parts of the system.
Example 7: Throwing a Ball from a Skateboard
A 60 kg person standing on a skateboard is initially stationary.
They throw a 2 kg ball right at 12 m/s.
Ignore the skateboard's mass for simplicity.
Initial momentum:
0
Ball momentum:
p = (2)(12)
p = +24 kg·m/s
Therefore, the person's momentum must be:
−24 kg·m/s
So:
60v = −24
v = −0.40 m/s
The person moves:
0.40 m/s to the left.
Why the Person Moves Slowly
The ball and person have equal-magnitude opposite momenta.
Ball:
24 kg·m/s
Person:
24 kg·m/s
But their masses are very different.
Because:
v = p/m
the large mass of the person means a much smaller recoil speed.
Example 8: Jumping from a Boat
A person and a small boat are initially stationary.
Person:
mass = 60 kg
Boat:
mass = 120 kg
The person jumps right at 3 m/s relative to the shore.
Ignoring external horizontal forces:
0 = (60)(3) + (120)v
0 = 180 + 120v
v = −1.5 m/s
The boat moves:
1.5 m/s left.
This is another example of recoil without a conventional explosion.
Relative Velocity Caution
In some advanced recoil problems, a velocity may be given relative to another moving object rather than relative to the ground.
For example:
"the person jumps at 3 m/s relative to the boat"
is not necessarily the same as:
"the person moves at 3 m/s relative to the shore."
Always identify the reference frame before using a velocity in the momentum equation.
Example 9: Moving System and Recoil
A 100 kg cart carrying a 5 kg package moves right at 4 m/s.
The package is launched forward at 10 m/s relative to the ground.
Afterward, find the cart's velocity.
Initial total mass:
105 kg
Initial momentum:
pᵢ = (105)(4)
pᵢ = 420 kg·m/s
After:
Package momentum:
(5)(10) = 50 kg·m/s
Cart momentum:
100v
Therefore:
420 = 50 + 100v
370 = 100v
v = 3.7 m/s
The cart continues moving right but slows from:
4.0 m/s to 3.7 m/s
because some forward momentum has been transferred to the package.
Energy in Explosions
Explosions involve both momentum and energy.
Momentum tells us how the motion of the fragments must balance.
Energy tells us where the kinetic energy came from.
For example:
Compressed spring:
elastic potential → kinetic
Rocket:
chemical → thermal + kinetic + other forms
The conservation laws work together but describe different aspects of the interaction.
Momentum Can Be Zero While Kinetic Energy Is Large
Suppose two equal masses move apart at equal speeds.
Object A:
momentum = +50 kg·m/s
Object B:
momentum = −50 kg·m/s
Total momentum:
0
Yet both objects are moving, so both have kinetic energy.
Therefore:
zero total momentum does not mean zero kinetic energy.
This is especially important when analyzing explosions from rest.
Example 10: Energy After an Explosion
A 2 kg cart and a 4 kg cart are initially stationary.
After a spring is released:
2 kg cart → +6 m/s
Momentum:
+12 kg·m/s
Therefore the 4 kg cart must have:
−12 kg·m/s
Its velocity is:
v = −12/4
v = −3 m/s
Now calculate kinetic energy.
2 kg cart:
KE = ½(2)(6²) = 36 J
4 kg cart:
KE = ½(4)(3²) = 18 J
Total kinetic energy:
54 J
That energy must have come from stored energy in the system.
Recoil and Impulse
Recoil can also be analyzed using impulse.
Impulse:
J = Δp
If two parts of a system interact:
Δp₁ = −Δp₂
Therefore:
J₁ = −J₂
The impulses are equal in magnitude and opposite in direction.
This is why the momentum changes balance.
Force and Recoil
Because:
F = Δp/Δt
a large momentum transfer over a short time produces a large force.
If the same momentum transfer occurs over a longer time, the average force is smaller.
This links recoil problems with the earlier topic of impulse.
Real-World Application: Spacecraft Maneuvering
Spacecraft need to change their velocity and orientation while in space.
Small thrusters can expel gas in one direction.
The spacecraft gains momentum in the opposite direction.
Different thrusters can be used to:
- accelerate
- decelerate
- rotate
- adjust orientation
- modify an orbit
Momentum conservation remains central to the process.
Reaction Wheels
Some spacecraft also use reaction wheels to change orientation without continuously expelling propellant.
An electric motor speeds up a wheel inside the spacecraft.
The wheel gains angular momentum in one direction.
The spacecraft rotates in the opposite direction.
This involves conservation of angular momentum, which is related to but different from the linear momentum studied here.
Real-World Application: Rocket Staging
Many launch vehicles use multiple stages.
As fuel is consumed and empty stages are released, the vehicle's mass decreases.
Reducing unnecessary mass allows the remaining propulsion system to accelerate the useful payload more effectively.
Real rocket calculations require variable-mass mechanics, but the underlying idea of momentum exchange between vehicle and exhaust remains fundamental.
Real-World Application: Water Rockets
A water rocket provides a useful classroom example of momentum-based propulsion.
Compressed air forces water downward out of the bottle.
The water gains downward momentum.
The rocket gains upward momentum.
Energy initially stored in the compressed air is converted into kinetic energy of the water and rocket.
This demonstrates:
- momentum conservation
- Newton's Third Law
- pressure
- energy transfer
- propulsion
in one system.
A Reliable Explosion and Recoil Strategy
Use the following method.
Step 1: Define the system.
Identify all relevant objects.
Step 2: Determine the initial momentum.
Was the system stationary or moving?
Step 3: Choose a positive direction.
For example:
right = positive.
Step 4: Write the momentum of each object.
Use:
p = mv
Step 5: Apply conservation of momentum.
Σp(before) = Σp(after)
Step 6: Include signs carefully.
Opposite directions require opposite signs.
Step 7: Solve for the unknown.
This may be:
- velocity
- momentum
- mass
Step 8: Interpret the sign.
A negative velocity means motion opposite to your chosen positive direction.
Step 9: Check total momentum.
Momentum before should equal momentum after.
Step 10: Consider energy.
If the objects gained kinetic energy, identify the likely energy source.
Worked Problem 1
A stationary 15 kg object separates into two pieces.
Piece A:
mass = 5 kg
velocity = +8 m/s
Piece B:
mass = 10 kg
velocity = unknown
Initial momentum:
0
After:
0 = (5)(8) + 10v
0 = 40 + 10v
v = −4 m/s
Therefore:
Piece B moves:
4 m/s in the opposite direction.
Worked Problem 2
Two carts initially at rest push apart.
Cart A:
mass = 2 kg
velocity = −10 m/s
Cart B:
mass = 5 kg
velocity = unknown
0 = (2)(−10) + 5v
0 = −20 + 5v
v = +4 m/s
Cart B moves:
4 m/s in the opposite direction.
Worked Problem 3
A stationary 80 kg person on a wheeled platform throws a 4 kg object right at 10 m/s.
Ignoring the platform mass:
Object momentum:
p = 4 × 10 = +40 kg·m/s
Person:
80v = −40
v = −0.5 m/s
The person recoils:
0.5 m/s left.
Worked Problem 4
A 20 kg moving object travels right at 6 m/s and separates into two 10 kg pieces.
One piece moves right at 9 m/s.
Find the other piece's velocity.
Initial momentum:
pᵢ = (20)(6)
pᵢ = 120 kg·m/s
After:
120 = (10)(9) + (10)v
120 = 90 + 10v
30 = 10v
v = 3 m/s
Both fragments move right.
This is possible because the system had forward momentum before separation.
Worked Problem 5
A stationary object separates into three fragments.
Fragment A:
+30 kg·m/s
Fragment B:
−18 kg·m/s
Fragment C:
unknown
Initial momentum:
0
Therefore:
0 = 30 − 18 + pC
pC = −12 kg·m/s
Worked Problem 6: Rocket Principle
A simplified propulsion system expels 2 kg of gas backward at 50 m/s relative to the chosen reference frame.
Gas momentum:
p = (2)(−50)
p = −100 kg·m/s
If the system initially had zero total momentum, the remaining vehicle must gain:
+100 kg·m/s
of momentum.
If the vehicle mass is 25 kg:
25v = 100
v = 4 m/s
This is a simplified model; real rocket motion involves continuously changing mass.
Worked Problem 7: Check the Prediction
Two carts start at rest.
Cart A:
mass = 3 kg
Cart B:
mass = 9 kg
Which cart should move faster after they push apart?
Because their momentum magnitudes must be equal:
3vA = 9vB
Therefore:
vA = 3vB
The 3 kg cart moves three times as fast as the 9 kg cart.
This can be predicted before any numerical velocities are known.
Common Mistakes
Mistake 1: Assuming explosions destroy momentum
Momentum is conserved in an isolated system.
An explosion redistributes momentum.
Mistake 2: Giving both objects positive velocities
If an initially stationary two-object system separates in opposite directions, one velocity must be positive and the other negative.
Mistake 3: Assuming equal momentum means equal velocity
Equal momentum magnitudes only produce equal speeds if the masses are equal.
Mistake 4: Assuming the heavier object recoils faster
For equal momentum magnitude:
v = p/m
The heavier object moves more slowly.
Mistake 5: Setting initial momentum to zero in every explosion
Only do this when the original system was stationary in the chosen reference frame.
Mistake 6: Thinking kinetic energy must remain zero because momentum is zero
Two objects can have equal and opposite momentum while both possess kinetic energy.
Mistake 7: Saying rockets push against air
Rockets expel propellant and gain momentum in the opposite direction.
They work in a vacuum.
Mistake 8: Confusing force with momentum
Force causes a change in momentum.
Momentum itself is:
p = mv
Mistake 9: Forgetting the energy source
When objects accelerate apart, their kinetic energy comes from stored energy such as elastic, chemical, electrical, pressure, or nuclear energy.
Mistake 10: Ignoring the reference frame
Velocities must be measured relative to the same reference frame before they are substituted into a momentum equation.
Did You Know?
One of the clearest demonstrations of momentum conservation can happen in almost complete silence in space.
A spacecraft can fire a thruster and accelerate even though there is essentially no surrounding air.
The spacecraft does not need anything outside itself to "push against."
Instead, it expels propellant.
The propellant carries momentum one way while the spacecraft gains momentum in the opposite direction.
The same principle can be demonstrated in a classroom with a balloon.
The scale changes enormously.
The physics does not.
Key Terms
Explosion: An interaction in which parts of a system move apart as stored energy is released.
Recoil: Motion of one part of a system caused by another part gaining momentum in the opposite direction.
Momentum: A vector quantity determined by mass and velocity.
Conservation of momentum: The principle that total momentum remains constant in an isolated system.
Isolated system: A system experiencing negligible net external impulse during the interaction.
Propulsion: Producing motion by transferring momentum to another mass.
Exhaust: Material expelled from a propulsion system.
Thrust: The force produced by a propulsion system.
Impulse: Change in momentum caused by a force acting over time.
Propellant: Material carried by a propulsion system and expelled to produce thrust.
Reference frame: The coordinate system relative to which motion is measured.
Key Equations
Momentum:
p = mv
Conservation of momentum:
Σp(before) = Σp(after)
Two-object system:
m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f
Explosion from rest:
0 = m₁v₁ + m₂v₂
Therefore:
m₁v₁ = −m₂v₂
Recoil velocity:
v₂ = −m₁v₁ / m₂
Impulse:
J = Δp
Average force:
F = Δp / Δt
Key Takeaways
- Explosions and recoil are applications of conservation of momentum.
- In physics, an explosion means that parts of a system move apart because stored energy is released.
- If an isolated system begins at rest, its total momentum is initially zero.
- Therefore, its total momentum must remain zero after the objects separate.
- In a simple two-object explosion from rest, the objects have equal-magnitude and opposite momenta.
- Equal momentum does not mean equal velocity.
- The lighter object moves faster when two objects have equal momentum magnitudes.
- Recoil occurs when one part of a system gains momentum in response to another part gaining momentum in the opposite direction.
- Newton's Third Law explains the equal and opposite forces that produce the momentum changes.
- Impulse provides another way to describe recoil because J = Δp.
- Explosions can increase kinetic energy because stored energy is transformed into kinetic energy.
- Zero total momentum does not mean zero kinetic energy.
- An explosion does not have to begin from rest.
- If the original system is moving, calculate its initial momentum before applying conservation.
- Fragments do not necessarily move in opposite directions if the original system was already moving.
- Momentum conservation also applies to systems containing more than two fragments.
- Balloon propulsion demonstrates recoil by expelling air backward.
- Rockets accelerate by expelling propellant backward.
- Rockets do not need atmospheric air to produce thrust and can operate in a vacuum.
- Jet engines, propellers, water jets, spacecraft thrusters, and biological jet propulsion all involve changes in fluid momentum.
- Real rocket calculations are more complex because rocket mass changes continuously as propellant is expelled.
- Velocities in a momentum calculation must be measured in the same reference frame.
- A reliable method for explosion and recoil problems is:
define the system → choose a direction → calculate initial momentum → write final momenta → apply conservation → solve → interpret the sign → check total momentum → identify the energy source.