What is momentum in the context of IB MYP Science?

In the IB MYP Science curriculum, momentum is defined as the product of an object's mass and its velocity. It is a vector quantity, meaning it has both magnitude and direction. Momentum is a key concept in understanding motion, forces, and energy, as it helps explain how objects interact and move in response to forces.

How is momentum calculated and what are its units?

Momentum is calculated using the formula: momentum (p) = mass (m) × velocity (v). The units of momentum are typically kilogram meters per second (kg·m/s) in the International System of Units (SI). Understanding this calculation is crucial for solving problems related to motion and forces in the IB MYP Science curriculum.

Why is momentum considered a conserved quantity?

Momentum is considered a conserved quantity because, in a closed system with no external forces, the total momentum before an interaction is equal to the total momentum after the interaction. This principle, known as the conservation of momentum, is fundamental in analyzing collisions and explosions in physics, making it an essential concept in the study of motion, forces, and energy in the IB MYP Science curriculum.

What is the difference between momentum and kinetic energy?

Momentum and kinetic energy are both related to the motion of objects, but they are different concepts. Momentum is the product of mass and velocity, while kinetic energy is the energy an object possesses due to its motion, calculated as 0.5 × mass × velocity squared. Momentum is a vector quantity, whereas kinetic energy is a scalar. Understanding these differences is important for students studying motion, forces, and energy in the IB MYP Science curriculum.

How does impulse relate to momentum in the study of motion?

Impulse is the change in momentum of an object when a force is applied over a period of time. It is calculated as the product of the force and the time duration over which it acts. In the IB MYP Science curriculum, understanding impulse helps explain how forces affect the motion of objects, as it directly relates to the change in momentum, providing insights into how objects accelerate or decelerate under various forces.

Why is "Adding momenta without considering direction." a common mistake in Momentum?

Why it happens: Most students treat numbers as positive scalars. How to avoid it: Assign + and − for opposite directions on a 1-D problem, then add. Always include direction in the final answer.

Why is "Applying conservation of momentum when external forces are present." a common mistake in Momentum?

Why it happens: Students see a collision and reach for the formula without checking the system. How to avoid it: Check that during the very brief interaction time, external forces (friction, gravity) have negligible effect compared to the interaction force.

Why is "Saying "airbags reduce the force because they slow the car down"." a common mistake in Momentum?

Why it happens: It feels intuitive — they cushion the impact. How to avoid it: Airbags don't slow the car — they extend the time over which the passenger's momentum changes. Same Δp, larger Δt, so smaller F.

Motion, forces and energyIB MYP Sciences (Years 1-5)

Momentum

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Momentum — IB MYP Sciences: p = mv, conservation of momentum and impulse

Momentum is the "oomph" carried by a moving object. This MYP Sciences note explains p = mv, why momentum is conserved in collisions and explosions, and how the F = Δp/Δt form of Newton's second law explains crumple zones, airbags and cricket bats.

What you’ll learn

Mapped to the IB MYP Sciences subject guide (2026 onwards).

MYP Sciences A — Define momentum and state its unit.
MYP Sciences A — Apply conservation of momentum to 1-D collisions and explosions.
MYP Sciences A — Use F = Δp / Δt to explain why crumple zones, airbags and helmets reduce injuries.
MYP Sciences D — Evaluate the role of momentum-based safety features in cars and bicycles.

What is momentum?

Momentum measures how hard it is to stop a moving object.

A lorry rolling at 1 m/s is harder to stop than a football at 1 m/s — because it has much more mass. A bullet at 400 m/s is harder to stop than a ball at 1 m/s — because it has much more velocity. Both effects combine into one quantity, momentum:

$p = m \, v$

where p is momentum (kg⋅m/s), m is mass (kg) and v is velocity (m/s).

Because velocity has direction, momentum is a vector — it has both size and direction. In MYP problems set on a straight line, we use + and − to show direction.

Object	Mass	Velocity	Momentum
Football	0.45 kg	25 m/s	11.25 kg⋅m/s
Cricket ball	0.16 kg	40 m/s	6.4 kg⋅m/s
Car	1 500 kg	14 m/s	21 000 kg⋅m/s

p = mv. Unit kg⋅m/s.
Vector — direction matters.
Larger mass or larger velocity → more momentum.

Conservation of momentum

In any closed system with no external resultant force, total momentum stays the same.

In a closed system (no external horizontal force), the total momentum before an interaction equals the total momentum after:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where u values are velocities before and v values are velocities after.

Before-and-after collision diagram: a 2 kg trolley at 3 m/s hits a stationary 1 kg trolley; after the collision they stick at 2 m/s

Worked example. A 2 kg trolley moving at 3 m/s hits a stationary 1 kg trolley and they stick together. Find the joint velocity.

Total before = (2 × 3) + (1 × 0) = 6 kg⋅m/s. Total after = (2 + 1) × v. So 3v = 6 → v = 2 m/s.

Explosions work the same way. A 50 g bullet fires from a 2.0 kg rifle. The bullet leaves at 400 m/s. What does the rifle do?

Before firing total momentum = 0 (both at rest). After firing: 0.050 × 400 + 2.0 × v = 0. → v = −10 m/s. The rifle recoils backwards at 10 m/s.

Total momentum before = total momentum after (closed system).
Applies equally to collisions and explosions.
Use + and − signs for opposite directions.

Force, impulse and safety design

Force equals rate of change of momentum: F = Δp / Δt. Longer impact times mean smaller forces.

Newton's second law is more general than F = ma. Its full form is:

$F = \frac{\Delta p}{\Delta t}$

The same change in momentum can produce a large force over a short time, or a small force over a long time.

This is the science behind every safety feature:

Crumple zones in cars deform during a crash, lengthening Δt and so reducing the force on passengers.
Airbags inflate to extend the time over which a passenger's head decelerates.
Cycle helmets crush their foam liner to extend the time of impact on your skull.
Bend the knees when landing from a jump — increases Δt, reduces force on joints.

The total change in momentum (= F × Δt) is also called the impulse. Impulse is measured in N⋅s, which equals kg⋅m/s.

F = Δp / Δt — change in momentum per unit time.
Impulse = F × Δt = change in momentum.
Longer Δt for the same Δp → smaller F. This is why all safety design extends impact time.

Sources: IB MYP Sciences guide (IBO, official subject guide). Last reviewed 2026-05-25.

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Step-by-step worked examples — Momentum

Step-by-step solutions to past-paper-style questions on momentum, written exactly the way a tutor would explain them at the board.

1Momentum of a runner
Getting started• p = mv
Question
A 70 kg sprinter runs at 9.0 m/s. Calculate their momentum.
Step-by-step solution
1. Step 1
  Use p = mv = 70 × 9.0.
2. Step 2
  = 630 kg⋅m/s in the direction of running.
Answer
630 kg⋅m/s in the direction of running.
2Sticking collision
Building confidence• conservation
Question
A 2.0 kg trolley moving at 3.0 m/s strikes a stationary 1.0 kg trolley and the two stick. Find their common velocity after the collision.
Step-by-step solution
1. Step 1
  Momentum before = (2.0)(3.0) + (1.0)(0) = 6.0 kg⋅m/s.
2. Step 2
  After: (2.0 + 1.0)v = 6.0 → v = 6.0 ÷ 3.0 = 2.0 m/s in the original direction.
Answer
2.0 m/s in the direction of the original motion.
3Recoil of a rifle
Stretch• conservation, explosion
Question
A 4.0 kg rifle fires a 0.020 kg bullet at 400 m/s. Calculate the recoil velocity of the rifle.
Step-by-step solution
1. Step 1
  Total momentum before = 0 (both at rest).
2. Step 2
  After: 0.020 × 400 + 4.0 × v = 0 → 8.0 + 4.0v = 0.
3. Step 3
  v = −2.0 m/s. Negative means opposite direction to the bullet.
Answer
2.0 m/s backwards (opposite to the bullet).
4Force from a change in momentum
Stretch• impulse, F = Δp/Δt
Question
A 0.15 kg cricket ball moving at 30 m/s is caught and stopped in 0.40 s. Calculate the average force on the catcher's hands.
Step-by-step solution
1. Step 1
  Δp = mΔv = 0.15 × (0 − 30) = −4.5 kg⋅m/s.
2. Step 2
  F = Δp ÷ Δt = −4.5 ÷ 0.40 = −11.25 N.
3. Step 3
  Magnitude = 11.25 N (directed against the ball's motion). Catching over a longer time would lower this force.
Answer
11.25 N against the ball's motion.

Key Formulae — Momentum

The formulae you need to memorise for momentum on the IB MYP Sciences paper, with every variable defined in plain English and a note on when to use it.

Momentum
$p = m × v$
$p$
momentum (kg⋅m/s)
$m$
mass (kg)
$v$
velocity (m/s, vector)
When to use
For any moving object.
Conservation of momentum
$m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂$
$u$
velocity before collision
$v$
velocity after collision
When to use
Closed system: no external resultant force during the interaction.
Force as rate of change of momentum
$F = Δp / Δt$
$F$
force (N)
$Δp$
change in momentum (kg⋅m/s)
$Δt$
time over which momentum changes (s)
When to use
Newton's 2nd law in its general form — useful when mass changes or when analysing impact times.

Key Definitions and Keywords — Momentum

Definitions to memorise and the exact keywords mark schemes credit for momentum answers — sharpened from recent examiner reports for the 2026 IB MYP Sciences sitting.

Momentum
Examiner keyword
Mass × velocity. Vector quantity; units kg⋅m/s.
Conservation of momentum
Examiner keyword
Total momentum of a closed system stays constant during any interaction (collision or explosion).
Impulse
Force × time of action = change in momentum. Units N⋅s = kg⋅m/s.
Closed system
A system with no external resultant force acting on it during the interaction of interest.

Common Mistakes and Misconceptions — Momentum

The traps other students keep falling into on momentum questions — taken from recent IB MYP Sciences examiner reports and mark schemes — and how to avoid them.

✕Adding momenta without considering direction.
Why it happens
Most students treat numbers as positive scalars.
How to avoid it
Assign + and − for opposite directions on a 1-D problem, then add. Always include direction in the final answer.
✕Applying conservation of momentum when external forces are present.
Why it happens
Students see a collision and reach for the formula without checking the system.
How to avoid it
Check that during the very brief interaction time, external forces (friction, gravity) have negligible effect compared to the interaction force.
✕Saying "airbags reduce the force because they slow the car down".
Why it happens
It feels intuitive — they cushion the impact.
How to avoid it
Airbags don't slow the car — they extend the time over which the passenger's momentum changes. Same Δp, larger Δt, so smaller F.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Momentum — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Object

Mass

Velocity

Momentum

Football

0.45 kg

25 m/s

11.25 kg⋅m/s

Cricket ball

0.16 kg

40 m/s

6.4 kg⋅m/s

Car

1 500 kg

14 m/s

21 000 kg⋅m/s

In a closed system (no external horizontal force), the total momentum before an interaction equals the total momentum after:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where u values are velocities before and v values are velocities after.

Before-and-after collision diagram: a 2 kg trolley at 3 m/s hits a stationary 1 kg trolley; after the collision they stick at 2 m/s

Worked example. A 2 kg trolley moving at 3 m/s hits a stationary 1 kg trolley and they stick together. Find the joint velocity.

Total before = (2 × 3) + (1 × 0) = 6 kg⋅m/s. Total after = (2 + 1) × v. So 3v = 6 → v = 2 m/s.

Explosions work the same way. A 50 g bullet fires from a 2.0 kg rifle. The bullet leaves at 400 m/s. What does the rifle do?

Before firing total momentum = 0 (both at rest). After firing: 0.050 × 400 + 2.0 × v = 0. → v = −10 m/s. The rifle recoils backwards at 10 m/s.

Momentum

Short Study Notes in the form of Slides

Short Study Notes — Momentum

Detailed Notes

What you’ll learn

1Momentum of a runner

2Sticking collision

3Recoil of a rifle

4Force from a change in momentum

Momentum

Conservation of momentum

Force as rate of change of momentum

Momentum

Conservation of momentum

Impulse

Closed system

✕Adding momenta without considering direction.

✕Applying conservation of momentum when external forces are present.

✕Saying "airbags reduce the force because they slow the car down".

Practice questions

Past paper style quiz

Assess your understanding

Momentum — frequently asked questions

Momentum

Short Study Notes in the form of Slides

Short Study Notes — Momentum

Detailed Notes

What you’ll learn

1Momentum of a runner

2Sticking collision

3Recoil of a rifle

4Force from a change in momentum

Momentum

Conservation of momentum

Force as rate of change of momentum

Momentum

Conservation of momentum

Impulse

Closed system

✕Adding momenta without considering direction.

✕Applying conservation of momentum when external forces are present.

✕Saying "airbags reduce the force because they slow the car down".

Practice questions

Past paper style quiz

Assess your understanding

Momentum — frequently asked questions