Momentum and Impulse
Momentum measures how hard something is to stop. It is conserved in every collision and explosion, which makes it one of the most powerful problem-solving tools in the syllabus — and the reason direction matters so much here.
1. Momentum
Momentum = mass × velocity p = m v
Unit: kg m/s (there is no special name).
Momentum is a VECTOR — it has direction.
Example: a 1500 kg car at 20 m/s.
- p = 1500 × 20 = 30 000 kg m/s
Mass must be in kilograms. Converting grams to kilograms was a recorded error.
Assign a positive direction and stick to it. Anything travelling the other way has a negative velocity, and therefore negative momentum. Forgetting to include the sign was a recorded error, and it is the main reason collision questions go wrong.
2. Conservation of momentum
The total momentum before a collision equals the total momentum after, provided no external force acts.
total momentum before = total momentum after
Method for any collision problem:
- Choose a positive direction
- Write the total momentum before (with signs)
- Write the total momentum after (with signs)
- Set them equal and solve
Objects that stick together
Example: a 2 kg trolley at 3 m/s hits a stationary 4 kg trolley and they move off together.
- Before: (2 × 3) + (4 × 0) = 6 kg m/s
- After: (2 + 4) × v = 6v
- 6v = 6 → v = 1 m/s
Add the masses when objects join. Their combined mass moves at the new velocity.
Objects moving towards each other
Example: 3 kg at 4 m/s right meets 2 kg at 5 m/s left, and they stick together.
- Taking right as positive: before = (3 × 4) + (2 × −5) = 12 − 10 = 2 kg m/s
- After: 5v = 2 → v = 0.4 m/s to the right
Subtract when objects move in opposite directions — or rather, include the negative sign and let the arithmetic handle it. Adding velocities that should have been subtracted was a specific recorded error.
A negative answer means the object moves the other way — that is a valid result, not a mistake.
Explosions
Before an explosion the total momentum is usually zero, so afterwards the two parts have equal and opposite momenta.
Example: a stationary 200 kg cannon fires a 2 kg ball at 150 m/s.
- Before = 0
- After: (2 × 150) + (200 × v) = 0
- 300 + 200v = 0 → v = −1.5 m/s — the cannon recoils at 1.5 m/s
3. Impulse
Impulse = force × time = CHANGE in momentum F t = Δp = m v − m u
Unit: N s — which is equivalent to kg m/s.
Impulse IS the change in momentum — they are the same quantity, not two different ones. Impulse is not simply “mass × velocity”; that is momentum itself. Calculating impulse as mv rather than mv − mu was a recorded error.
Example: a 0.5 kg ball hits a wall at 8 m/s and rebounds at 6 m/s.
- Taking towards the wall as positive: u = +8, v = −6
- Δp = m(v − u) = 0.5 × (−6 − 8) = 0.5 × (−14) = −7 kg m/s
- The magnitude of the impulse is 7 N s
A rebound reverses the sign of the velocity, so the change in momentum is larger than you might expect — this is why bouncing exerts more force than stopping dead.
4. Force as rate of change of momentum
F = Δp / t — force equals the rate of change of momentum
Example: momentum changes by 400 kg m/s in 0.2 s.
- F = 400 ÷ 0.2 = 2000 N
Divide the change in momentum by the TIME, and use the change, not the final value. Confusing “final time” with “final momentum”, and dividing incorrectly, were both recorded.
Why this matters — safety features
This equation explains every safety device in the syllabus.
For a given change in momentum, increasing the time of the collision reduces the force.
| Feature | How it works |
|---|---|
| Seat belts | stretch slightly, increasing the time to stop, so less force on the body |
| Crumple zones | deform, extending the collision time and reducing the force |
| Air bags | increase the stopping time and spread the force over a larger area |
| Crash mats / helmets | compress, increasing the time |
Say “increases the time, so reduces the force”. A recorded error explained seat belts as “preventing the velocity of the passenger” — the mark scheme wants the time–force argument. The passenger’s change in momentum is the same; the belt changes how long it takes.
Structure a three-mark answer as three statements: (1) the change in momentum is the same, (2) the feature increases the time taken, (3) so from F = Δp/t the force is reduced. Tutors flagged exactly this three-part structure.
5. Mistakes that cost marks
Ignoring direction — no negative velocities.
Adding velocities that act in opposite directions.
Leaving mass in grams.
Calculating impulse as mv instead of the change in momentum.
Using the final momentum instead of the change in F = Δp/t.
Forgetting to add masses when objects stick together.
Treating a negative answer as an error.
Explaining safety features without the time–force argument.
Wrong units — kg m/s for momentum, N s for impulse.
Frequently asked questions
What is momentum? Mass × velocity (p = mv), measured in kg m/s. It is a vector.
What is the principle of conservation of momentum? Total momentum before = after, when no external force acts.
How do I handle objects moving in opposite directions? Choose a positive direction and give the other velocity a negative sign.
What happens when objects stick together? Add their masses; they move off with a common velocity.
What is an explosion in momentum terms? Total momentum before is zero, so the fragments have equal and opposite momenta.
What is impulse? Force × time, which equals the change in momentum. Unit: N s.
What is the equation for force in terms of momentum? F = Δp / t — the rate of change of momentum.
How do seat belts reduce injury? They increase the time taken to stop, so the force is reduced.
Why does a rebound involve a larger change in momentum? Because the velocity reverses direction, so the change is v − (−u).
What are the units of momentum? kg m/s.
Quick revision checklist
- I know p = mv and its unit
- I know momentum is a vector
- I assign a positive direction and use signs
- I can apply conservation of momentum to collisions
- I add masses when objects stick together
- I can handle head-on collisions with negative velocities
- I can solve explosion problems from zero total momentum
- I know impulse = Ft = Δp
- I use the change in momentum, including rebounds
- I know F = Δp/t
- I can explain safety features with the time–force argument
- I can structure that explanation in three statements
- I convert grams to kilograms
- I give the right units
These notes cover momentum and impulse in the Cambridge IGCSE Physics (0625) syllabus and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Physics lessons, with particular attention to the errors students make most often and the wording examiners reward. Always check the current syllabus and formula list for your own exam series.
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