What is momentum in Physics and how is it calculated?

In Physics, momentum is a measure of the quantity of motion an object has. It is calculated as the product of an object's mass and velocity. The formula for momentum (p) is p = m × v, where m is the mass in kilograms and v is the velocity in meters per second. Momentum is a vector quantity, meaning it has both magnitude and direction.

Why is momentum considered a vector quantity?

Momentum is considered a vector quantity because it depends on both the magnitude and direction of an object's motion. This means that when calculating momentum, both the speed and the direction of the object's velocity must be taken into account. In Physics, especially at the Cambridge IGCSE level, understanding vector quantities is crucial for analyzing motion and forces.

How does the principle of conservation of momentum apply in collisions?

The principle of conservation of momentum states that in a closed system, the total momentum before a collision is equal to the total momentum after the collision, provided no external forces act on the system. This principle is fundamental in Physics and is used to solve problems involving collisions, where the momentum of the objects involved is conserved. This concept is particularly important in the Cambridge IGCSE curriculum when studying motion and forces.

What is the difference between elastic and inelastic collisions in terms of momentum?

In an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy of the system remain the same before and after the collision. In contrast, in an inelastic collision, momentum is conserved, but kinetic energy is not. Some kinetic energy is transformed into other forms of energy, such as heat or sound. Understanding these differences is essential for Cambridge IGCSE Physics students studying motion and energy.

How does impulse relate to momentum in Physics?

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. The formula for impulse (J) is J = F × Δt, where F is the force in newtons and Δt is the time in seconds. Impulse is equal to the change in momentum, which is a key concept in analyzing how forces affect motion in Physics, particularly in the Cambridge IGCSE curriculum.

Why is "Adding instead of subtracting after a rebound" a common mistake in Momentum?

Why it happens: Forgetting that velocity reverses direction. How to avoid it: Pick a positive direction at the start. Reversed velocity gets a negative sign. Source: 0625/42 — recurring

Why is "Quoting impulse in kg only" a common mistake in Momentum?

Why it happens: Forgetting the velocity factor. How to avoid it: Impulse has units N s = kg m/s. Always include both parts.

Why is "Using separate final velocities when the objects stick together" a common mistake in Momentum?

Why it happens: Misreading the question. How to avoid it: Stuck after collision → both move with the SAME final velocity. Bounce → separate finals.

Why is "Forgetting to add masses in 'stick together' collisions" a common mistake in Momentum?

Why it happens: Treating the new combined object as having only the moving mass. How to avoid it: After sticking, total mass = m_1 + m_2.

Cambridge IGCSE Physics (0625)

Momentum

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Short Notes - Momentum

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Detailed Study Notes

Detailed notes on Motion, Forces and Energy for Cambridge IGCSE Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Momentum — Cambridge IGCSE 0625 Physics Extended (2026)

$p = mv$ as a vector. Conservation of momentum in collisions and explosions, plus Newton's second law in its momentum form: $F = \Delta p / \Delta t$ .

At a glance

Momentum $p = mv$ . Vector. Units: $\text{kg m/s}$ .
Conservation: total momentum before = total momentum after (closed system).
Force = rate of change of momentum: $F = \Delta p / \Delta t$ .
Impulse = $F \Delta t = \Delta p$ .
Direction matters — choose a positive direction and stick to it.
Crumple zones, airbags, helmets all increase $\Delta t$ to reduce $F$ .

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

1.6 — Define momentum as $p = mv$ .
1.6 — State and apply the principle of conservation of momentum to collisions in one dimension.
1.6 — Apply $F = \Delta p / \Delta t$ to calculate force in a collision.

Defining momentum

$p = mv$ . Vector. Sign matches direction.

Momentum. $p = mv.$

$p$ : momentum ( $\text{kg m/s}$ ).
$m$ : mass ( $\text{kg}$ ).
$v$ : velocity ( $\text{m/s}$ ).

Vector. Direction matters — momentum to the LEFT is the negative of momentum to the RIGHT.

Worked. A $0.5\,\text{kg}$ ball at $4\,\text{m/s}$ .

$p = 0.5 \times 4 = 2\,\text{kg m/s}$ .

Worked. Same ball, now at $4\,\text{m/s}$ in the OPPOSITE direction.

$p = -2\,\text{kg m/s}$ (taking original direction as positive).

$p = mv$ .
Vector — choose +ve direction and stick to it.
Units: $\text{kg m/s}$ .

Conservation of momentum

In any closed system (no external force), total momentum before = total momentum after.

Principle. $p_{\text{before, total}} = p_{\text{after, total}}.$

Applies to collisions and explosions.

Worked (collision). A $2\,\text{kg}$ trolley at $3\,\text{m/s}$ collides with a stationary $1\,\text{kg}$ trolley. They stick. Find their common velocity.

Before: $p = 2 \times 3 + 1 \times 0 = 6\,\text{kg m/s}$ .
After: $(2 + 1) \times v = 6 \Rightarrow v = 2\,\text{m/s}$ .

Worked (explosion). A stationary cannon ( $500\,\text{kg}$ ) fires a $5\,\text{kg}$ shell at $200\,\text{m/s}$ . Find recoil velocity.

Before: $p = 0$ .
After: $5 \times 200 + 500 \times v_{\text{cannon}} = 0$ .
$v_{\text{cannon}} = -1000/500 = -2\,\text{m/s}$ (recoils opposite to shell).

Tip. Always set up "before" and "after" rows and equate. Direction signs matter.

In a closed system, the total momentum is the same before and after the collision.

Total $p$ conserved in closed system.
Set up before-after table.
Watch direction signs.
Works for collisions AND explosions.

Force as rate of change of momentum

$F = \Delta p / \Delta t$ . The original form of Newton's second law.

Newton's second law (momentum form). $F = \dfrac{\Delta p}{\Delta t} = \dfrac{m(v - u)}{\Delta t}.$

This is identical to $F = ma$ when mass is constant.

Worked. A $0.05\,\text{kg}$ tennis ball at $20\,\text{m/s}$ is hit back at $25\,\text{m/s}$ . Contact time $0.005\,\text{s}$ .

$\Delta p = m(v - u) = 0.05 \times (-25 - 20) = -2.25\,\text{kg m/s}$ (taking incoming direction as positive).
$|F| = 2.25 / 0.005 = 450\,\text{N}$ .

Impulse. $F \Delta t = \Delta p$ . Same idea, rearranged.

Safety applications. Crumple zones, airbags, helmets, sand pits all INCREASE the time over which momentum changes, REDUCING the force on the body.

The same change in momentum spread over a longer time means a much smaller peak force.

$F = \Delta p / \Delta t$ .
Impulse $= F \Delta t = \Delta p$ .
Increase $\Delta t$ → reduce $F$ for same $\Delta p$ .
Safety devices apply this idea.

Quick recap

$p = mv$ . Vector.
Conservation: total $p$ before = after.
$F = \Delta p / \Delta t$ .
Impulse $= F \Delta t = \Delta p$ .
Safety devices increase $\Delta t$ to reduce $F$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Momentum — product of mass and velocity, vector.
Conservation of momentum — total momentum is constant in a closed system.
Impulse — product of force and time of application, equals change in momentum.

How it’s examined

Momentum appears every Paper 4 (5-7 marks) — a collision-conservation calculation followed by a $F = \Delta p / \Delta t$ part on a safety device. Examiner reports flag two errors: missing negative sign for opposite direction, and using mass instead of momentum in the conservation equation.

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (1.6); 0625/42 Oct/Nov 2024 — Q7 (collision); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

Take this whole topic with you

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.

Question type:

Question patterns to master — Momentum

Almost every momentum exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single calculate or find instruction using one momentum formula — $p = mv$ , impulse $= Ft$ , or conservation $m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$ .

How to approach it

Choose a positive direction first, write the formula, substitute with signed velocities, then give the answer with its unit (kg m/s for momentum, N s for impulse).

Common trap

Examiner reports flag candidates forgetting to reverse the sign of a velocity after a rebound — momentum is a vector, so direction matters.

Multi-step problem

Recognise it by

Several stages are chained — two scenarios compared, or conservation of momentum followed by a kinetic-energy check.

How to approach it

Work each stage in turn with a clear sub-answer, keeping the same positive direction throughout; only combine the results at the end.

Common trap

Assuming kinetic energy is conserved. Momentum is always conserved in a closed system, but kinetic energy is conserved only in an elastic collision.

1Calculate momentum
CoreDirect calculation• p=mv
Question
A $1500\,\text{kg}$ car moves at $20\,\text{m/s}$ . Find its momentum.
Step-by-step solution
1. Step 1
  $p = mv$ .
  $p = 1500 \times 20 = 30{,}000\,\text{kg\,m/s}$
Answer
$30{,}000\,\text{kg\,m/s}$
2Impulse and change in momentum
ExtendedDirect calculation• Adapted from 0625/42 May/Jun 2024 Q6• impulse
Question
A $0.4\,\text{kg}$ ball travels at $5\,\text{m/s}$ and rebounds elastically off a wall at $5\,\text{m/s}$ in the opposite direction. Find the impulse.
Step-by-step solution
1. Step 1
  Take initial direction as positive.
  $\Delta p = mv - mu = 0.4(-5) - 0.4(5) = -4\,\text{kg\,m/s}$
2. Step 2
  Magnitude of impulse.
  $|\text{impulse}| = 4\,\text{kg\,m/s} = 4\,\text{N\,s}$
Answer
$4\,\text{N\,s}$ (directed away from the wall)
Examiner tip
Don't forget the sign change — initial and final velocities are in opposite directions.
3Conservation in a collision
ExtendedDirect calculation• conservation
Question
A $2\,\text{kg}$ trolley moving at $3\,\text{m/s}$ collides with a stationary $1\,\text{kg}$ trolley. They stick. Find the velocity after.
Step-by-step solution
1. Step 1
  Total momentum before = total after.
  $(2)(3) + (1)(0) = (2 + 1) v$
2. Step 2
  Solve.
  $v = \dfrac{6}{3} = 2\,\text{m/s}$
Answer
$2\,\text{m/s}$ in the same direction
4Average force from change in momentum
ExtendedDirect calculation• F=Δp/t
Question
A $0.05\,\text{kg}$ tennis ball arrives at $30\,\text{m/s}$ and leaves at $40\,\text{m/s}$ in the opposite direction after $0.01\,\text{s}$ of contact. Find the average force.
Step-by-step solution
1. Step 1
  Change in momentum.
  $\Delta p = 0.05(-40) - 0.05(30) = -3.5\,\text{kg\,m/s}$
2. Step 2
  Average force.
  $F = \dfrac{|\Delta p|}{t} = \dfrac{3.5}{0.01} = 350\,\text{N}$
Answer
$350\,\text{N}$
5Momentum from mass and speed
CoreDirect calculation• p=mv
Question
Compare the momenta of (a) a $50\,\text{kg}$ runner at $8\,\text{m/s}$ and (b) a $4000\,\text{kg}$ truck at $0.10\,\text{m/s}$ .
Step-by-step solution
1. Step 1
  Runner.
  $p_{r} = 50 \times 8 = 400\,\text{kg\,m/s}$
2. Step 2
  Truck.
  $p_{t} = 4000 \times 0.10 = 400\,\text{kg\,m/s}$
Answer
Both have momentum $400\,\text{kg\,m/s}$ — equal, despite very different masses and speeds.
6Change in momentum: bouncing vs sticking
ExtendedMulti-step problem• Adapted from 0625/22 May/Jun 2024 Q12• change in p
Question
A $0.20\,\text{kg}$ ball arrives at a wall at $6\,\text{m/s}$ . Find the change in its momentum if it (a) sticks to the wall, (b) rebounds elastically at $6\,\text{m/s}$ .
Step-by-step solution
1. Step 1
  Take initial direction toward the wall as positive.
2. Step 2
  (a) Sticks: final $v = 0$ .
  $\Delta p_{a} = 0.20(0) - 0.20(6) = -1.2\,\text{kg\,m/s}$
3. Step 3
  (b) Rebounds elastically: final $v = -6$ .
  $\Delta p_{b} = 0.20(-6) - 0.20(6) = -2.4\,\text{kg\,m/s}$
Answer
(a) $|\Delta p| = 1.2\,\text{kg\,m/s}$ . (b) $|\Delta p| = 2.4\,\text{kg\,m/s}$ — twice as much.
Examiner tip
The examiner report flags candidates often quote $\Delta p = 0$ for the elastic case because 'the speed is the same'. The DIRECTION changes, so the change in momentum is the largest — that's why bouncing is harder on the wall than sticking.
7Recoil — conservation in an 'explosion'
ExtendedDirect calculation• recoil, conservation
Question
A $4.0\,\text{kg}$ rifle fires a $0.020\,\text{kg}$ bullet at $300\,\text{m/s}$ . Find the recoil velocity of the rifle.
Step-by-step solution
1. Step 1
  Before firing, total momentum is zero (both at rest). By conservation, total after is also zero.
  $0 = m_{r} v_{r} + m_{b} v_{b}$
2. Step 2
  Solve for $v_{r}$ .
  $v_{r} = -\dfrac{m_{b} v_{b}}{m_{r}} = -\dfrac{0.020 \times 300}{4.0}$
3. Step 3
  Evaluate.
  $v_{r} = -1.5\,\text{m/s}$
Answer
$1.5\,\text{m/s}$ in the opposite direction to the bullet.
Examiner tip
The 2024 mark scheme awards method marks for explicitly stating that initial total momentum is zero — that is the entry point to the conservation argument.
8Impulse — why airbags reduce force
ExtendedMulti-step problem• impulse, concept
Question
A $60\,\text{kg}$ driver moving at $15\,\text{m/s}$ is brought to rest in a crash. Find the average force (a) over $0.05\,\text{s}$ (no airbag, hits steering wheel) and (b) over $0.30\,\text{s}$ (with airbag).
Step-by-step solution
1. Step 1
  Change in momentum is the same in both cases.
  $\Delta p = 60 \times 15 = 900\,\text{kg\,m/s}$
2. Step 2
  (a) No airbag.
  $F_{a} = \dfrac{900}{0.05} = 18{,}000\,\text{N}$
3. Step 3
  (b) With airbag.
  $F_{b} = \dfrac{900}{0.30} = 3000\,\text{N}$
Answer
$F_{a} = 18{,}000\,\text{N}$ ; $F_{b} = 3000\,\text{N}$ . The airbag reduces the force by a factor of $6$ by extending the stopping time.
Examiner tip
The examiner report flags candidates often say the airbag 'absorbs momentum'. The momentum CHANGE is the same; the airbag spreads it over a longer time, so $F = \Delta p / t$ is smaller.
9Two-car head-on collision — solve for unknown final velocity
ExtendedDirect calculation• conservation
Question
A $1200\,\text{kg}$ car moving east at $8\,\text{m/s}$ collides head-on with a $800\,\text{kg}$ car moving west at $5\,\text{m/s}$ . After the collision the heavier car continues east at $2\,\text{m/s}$ . Find the final velocity of the lighter car.
Step-by-step solution
1. Step 1
  Take east as positive. Apply conservation of momentum.
  $m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$
2. Step 2
  Substitute (west = negative).
  $1200(8) + 800(-5) = 1200(2) + 800\,v_{2}$
3. Step 3
  Simplify left and right sides.
  $9600 - 4000 = 2400 + 800\,v_{2}$
4. Step 4
  Solve.
  $v_{2} = \dfrac{5600 - 2400}{800} = 4\,\text{m/s east}$
Answer
$4\,\text{m/s}$ eastward — the lighter car is knocked backwards (it was originally going west).
Examiner tip
The examiner report flags candidates often drop the sign on the lighter car's initial velocity and get an unphysical answer. Always pick a positive direction at the start.
10A* — momentum AND energy together (inelastic collision)
ChallengeMulti-step problem• synoptic, KE
Question
A $3.0\,\text{kg}$ trolley moving at $4.0\,\text{m/s}$ collides with a stationary $1.0\,\text{kg}$ trolley and they stick. Find (a) the final velocity, (b) the kinetic energy before and after, and (c) state whether the collision is elastic or inelastic.
Step-by-step solution
1. Step 1
  Conservation of momentum.
  $(3.0)(4.0) + (1.0)(0) = (3.0 + 1.0) v$
2. Step 2
  Solve.
  $v = \dfrac{12}{4.0} = 3.0\,\text{m/s}$
3. Step 3
  KE before (only the moving trolley).
  $E_{k,i} = \tfrac{1}{2}(3.0)(4.0)^{2} = 24\,\text{J}$
4. Step 4
  KE after.
  $E_{k,f} = \tfrac{1}{2}(4.0)(3.0)^{2} = 18\,\text{J}$
5. Step 5
  KE decreased by $6\,\text{J}$ even though momentum is conserved → inelastic collision (energy transferred to heat/sound/deformation).
Answer
(a) $v = 3.0\,\text{m/s}$ . (b) KE before $= 24\,\text{J}$ , after $= 18\,\text{J}$ . (c) Inelastic — $6\,\text{J}$ of KE is transferred to other stores.
Examiner tip
The 2024 mark scheme awards an extra mark for stating that momentum is ALWAYS conserved in a closed system, but KE is only conserved in ELASTIC collisions.

Model Answers — Momentum

High-scoring sample answers for momentum on the Cambridge IGCSE 0625 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2/4 short-answer style1 mark
Define momentum and state its SI unit.
Model answer
Momentum is mass × velocity ( $p = mv$ ). It is a vector quantity, and its SI unit is the kilogram metre per second, $\text{kg}\,\text{m/s}$ (equivalently $\text{N}\,\text{s}$ ).
Why this scores
One mark for 'mass × velocity' together with the unit. Because velocity is a vector, momentum is a vector — useful to state when the next part involves a collision.
Question 2
Paper 2/4 style2 marks
Calculate the momentum of a $900\,\text{kg}$ car travelling at $15\,\text{m/s}$ .
Model answer
$p = mv = 900 \times 15 = \mathbf{13\,500\,\text{kg}\,\text{m/s}}.$
Why this scores
One mark for substitution into $p=mv$ , one for the value with the correct unit $\text{kg}\,\text{m/s}$ .
Question 3
Paper 4 structured style3 marks
A $2.0\,\text{kg}$ trolley moving at $3.0\,\text{m/s}$ collides with a stationary $1.0\,\text{kg}$ trolley and they stick together. State the principle of conservation of momentum and calculate their common velocity after the collision.
Model answer
Principle: in a closed system (no external resultant force), the total momentum before a collision equals the total momentum after.

Momentum before $= (2.0 \times 3.0) + (1.0 \times 0) = 6.0\,\text{kg}\,\text{m/s}$ . After sticking, the combined mass is $3.0\,\text{kg}$ moving at $v$ :

$6.0 = (2.0 + 1.0)\,v \implies v = \dfrac{6.0}{3.0} = \mathbf{2.0\,\text{m/s}}$

in the same direction as the original motion.
Why this scores
Three marks: statement of conservation (1), correct total momentum before with combined mass after (1), and $v = 2.0\,\text{m/s}$ (1). 'Stick together' means a single combined mass of $3.0\,\text{kg}$ — forgetting to add the masses is the usual slip.
Question 4
Paper 4 structured style4 marks
A $0.16\,\text{kg}$ cricket ball travelling at $25\,\text{m/s}$ is caught and brought to rest in $0.10\,\text{s}$ . (a) Calculate the change in momentum. (b) Calculate the average force the fielder's hands exert on the ball.
Model answer
(a) Change in momentum $= mv - mu = (0.16 \times 0) - (0.16 \times 25) = -4.0\,\text{kg}\,\text{m/s}$ , i.e. a magnitude of $\mathbf{4.0\,\text{kg}\,\text{m/s}}$ (directed opposite to the ball's motion).

(b) Average force $= \dfrac{\text{change in momentum}}{\text{time}} = \dfrac{4.0}{0.10} = \mathbf{40\,\text{N}}$ .
Why this scores
Four marks: change in momentum magnitude $4.0\,\text{kg}\,\text{m/s}$ (1) with a sign/direction comment (1); use of $F = \Delta p / t$ (1); $40\,\text{N}$ (1). This links to why a fielder 'gives' with the catch — a longer time means a smaller force.
Question 5
Paper 4 application style5 marks
A car has a $70\,\text{kg}$ driver moving at $20\,\text{m/s}$ . In a crash the driver is brought to rest. Explain, with a calculation, why an airbag that increases the stopping time from $0.10\,\text{s}$ to $0.50\,\text{s}$ reduces the force on the driver.
Model answer
The driver's change in momentum is the same in both cases, because the driver goes from $20\,\text{m/s}$ to rest regardless of how they are stopped:

$\Delta p = 70 \times 20 = 1400\,\text{kg}\,\text{m/s}.$

The force is $F = \dfrac{\Delta p}{t}$ , so spreading the same momentum change over a longer time gives a smaller force:

Without airbag: $F = \dfrac{1400}{0.10} = \mathbf{14\,000\,\text{N}}$

With airbag: $F = \dfrac{1400}{0.50} = \mathbf{2800\,\text{N}}$

The airbag extends the stopping time, so for the same change in momentum the average force is reduced (here by a factor of $5$ ), reducing the risk of injury.
Why this scores
Five marks: change in momentum is the same (1) and its value (1); $F = \Delta p / t$ (1); both force values (1); conclusion that a longer time gives a smaller force (1). The airbag does not 'absorb the momentum' — the momentum change is unchanged; only the time over which it happens increases.
Question 6
Paper 4 multi-part structured style6 marks
A $2.0\,\text{kg}$ trolley moving at $6.0\,\text{m/s}$ collides with a stationary $4.0\,\text{kg}$ trolley and they move off together. (a) Calculate their common velocity. (b) Calculate the total kinetic energy before and after the collision. (c) State, with a reason, whether the collision is elastic or inelastic.
Model answer
(a) Conservation of momentum: $(2.0 \times 6.0) + (4.0 \times 0) = (2.0 + 4.0)\,v$ , so $12 = 6.0\,v$ and $v = \mathbf{2.0\,\text{m/s}}$ .

(b) KE before $= \tfrac{1}{2}(2.0)(6.0)^{2} = \mathbf{36\,\text{J}}$ . KE after $= \tfrac{1}{2}(6.0)(2.0)^{2} = \mathbf{12\,\text{J}}$ .

(c) The collision is inelastic: kinetic energy has fallen from $36\,\text{J}$ to $12\,\text{J}$ — $24\,\text{J}$ has been transferred to other stores (heat, sound and deformation). Momentum is conserved in every collision, but kinetic energy is conserved only in an elastic collision.
Why this scores
Six marks: common velocity by conservation (1); KE before (1); KE after (1); identifying it as inelastic (1); KE is not conserved / energy transferred to heat etc. (1); the general statement that momentum is always conserved but KE only in elastic collisions (1).

Key Formulae — Momentum

The formulae you need to memorise for momentum on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Momentum
$p = mv$
$p$
momentum (kg m/s)
$m$
mass (kg)
$v$
velocity (m/s)
When to use
Any moving object; vector quantity.
Impulse / change in momentum
$\text{Impulse} = F t = \Delta p = m v - m u$
$F$
average force in N
$t$
contact time in s
When to use
Collisions where you know the contact time, or to relate force to velocity change.
Conservation of momentum
$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$
When to use
Collisions/explosions in a closed system (no external forces).

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 0625 sitting.

Momentum
Examiner keyword
Mass × velocity. Vector; SI unit kg m/s.
Impulse
Examiner keyword
The change in momentum produced by a force. Equal to $F \times t$ .
Principle of conservation of momentum
Examiner keyword
In a closed system the total momentum before equals the total momentum after.

Common Mistakes and Misconceptions — Momentum

The traps other students keep falling into on momentum questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Adding instead of subtracting after a rebound
0625/42 — recurring
Why it happens
Forgetting that velocity reverses direction.
How to avoid it
Pick a positive direction at the start. Reversed velocity gets a negative sign.
✕Quoting impulse in kg only
Why it happens
Forgetting the velocity factor.
How to avoid it
Impulse has units N s = kg m/s. Always include both parts.
✕Using separate final velocities when the objects stick together
Why it happens
Misreading the question.
How to avoid it
Stuck after collision → both move with the SAME final velocity. Bounce → separate finals.
✕Forgetting to add masses in 'stick together' collisions
Why it happens
Treating the new combined object as having only the moving mass.
How to avoid it
After sticking, total mass = $m_{1} + m_{2}$ .

Momentum — frequently asked questions

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

Momentum

Short Notes - Momentum

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Momentum

1Calculate momentum

2Impulse and change in momentum

3Conservation in a collision

4Average force from change in momentum

5Momentum from mass and speed

6Change in momentum: bouncing vs sticking

7Recoil — conservation in an 'explosion'

8Impulse — why airbags reduce force

9Two-car head-on collision — solve for unknown final velocity

10A* — momentum AND energy together (inelastic collision)

Momentum

Impulse / change in momentum

Conservation of momentum

Momentum

Impulse

Principle of conservation of momentum

✕Adding instead of subtracting after a rebound

✕Quoting impulse in kg only

✕Using separate final velocities when the objects stick together

✕Forgetting to add masses in 'stick together' collisions

Momentum — frequently asked questions