What is the difference between momentum and impulse in Physics?

In Physics, momentum is defined as the product of an object's mass and its velocity, represented by the formula p = mv. It is a vector quantity, meaning it has both magnitude and direction. Impulse, on the other hand, is the change in momentum of an object when a force is applied over a period of time. It is calculated using the formula J = FΔt, where J is impulse, F is the force applied, and Δt is the time duration. Understanding these concepts is crucial for Edexcel International A Levels students studying Further Mechanics, Fields and Particles.

How is the principle of conservation of momentum applied 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 analyzing collisions in Physics, especially for Edexcel International A Levels students. In elastic collisions, both momentum and kinetic energy are conserved, while in inelastic collisions, only momentum is conserved.

How do you calculate impulse from a force-time graph?

Impulse can be calculated from a force-time graph by finding the area under the curve. This area represents the product of force and time, which is the impulse. For Edexcel International A Levels students, understanding how to interpret these graphs is essential in the Further Mechanics, Fields and Particles section. The impulse gives the change in momentum of the object subjected to the force.

Why is impulse important in real-world applications?

Impulse is crucial in real-world applications because it helps us understand how forces affect the motion of objects over time. For example, in car safety design, crumple zones are engineered to increase the time over which a collision occurs, thereby reducing the force experienced by passengers. This concept is vital for Edexcel International A Levels students studying Physics, as it connects theoretical knowledge with practical applications in Further Mechanics, Fields and Particles.

What role does impulse play in sports physics?

In sports physics, impulse is used to analyze and improve performance by understanding how forces are applied over time. For instance, in activities like hitting a baseball or kicking a soccer ball, the impulse determines the change in momentum and thus the speed and direction of the ball. This understanding is beneficial for Edexcel International A Levels students, as it illustrates the application of momentum and impulse concepts in real-life scenarios within the Further Mechanics, Fields and Particles curriculum.

Why is "Ignoring direction (signs) when calculating a change of momentum or applying conservation" a common mistake in Momentum and Impulse?

Why it happens: Momentum is treated as a scalar, so velocities in opposite directions are added with the same sign — especially when a ball rebounds. How to avoid it: Momentum is a **vector**. Choose a positive direction and give opposite velocities **opposite signs**. When an object reverses, v = v - (-u), so the speeds **add** — e.g. 22 - (-9) = 31, not 13. Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Assuming kinetic energy is conserved in every collision (or that momentum conservation proves a collision is elastic)" a common mistake in Momentum and Impulse?

Why it happens: Students remember 'momentum is conserved' and wrongly extend it to kinetic energy, or think conserved momentum implies an elastic collision. How to avoid it: Momentum is conserved in **all** collisions. Kinetic energy is conserved **only** in elastic collisions. To classify a collision you MUST compare total KE before and after — a fall in KE means it is inelastic (energy → heat/sound/deformation). Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Misusing $E_k = p^2/2m$ — forgetting to square the momentum, or dividing by $m$ instead of $2m$" a common mistake in Momentum and Impulse?

Why it happens: The formula is confused with p = mv, and the factor of 2 and the square are dropped under time pressure. How to avoid it: Write the equation as E_k = p^2/2m and substitute carefully: **square p first**, then divide by 2m. Check units — (kg m s^-1)^2kg = kg m^2 s^-2 = J. Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Treating a two-dimensional collision as one-dimensional — adding speeds instead of resolving momentum into components" a common mistake in Momentum and Impulse?

Why it happens: Students apply the familiar 1D 'before = after' equation to angled velocities without resolving, so the perpendicular momentum is ignored. How to avoid it: In 2D, momentum is conserved **separately** in two perpendicular directions. Resolve every velocity into components (v and v), write **one conservation equation per axis**, and remember the perpendicular components must sum to the (usually zero) perpendicular momentum before. Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Giving the wrong unit for impulse or change of momentum, or confusing force with impulse" a common mistake in Momentum and Impulse?

Why it happens: Impulse (N s) and force (N) look similar, and momentum's unit kg m s^-1 is unfamiliar. How to avoid it: Impulse and change of momentum share the unit N s = kg m s^-1; a **force** is in newtons (N) alone. When a question asks for impulse, do not stop at the force — multiply by (or read the area over) the time, and quote N s. Source: Edexcel IAL Physics Unit 4 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Momentum and Impulse

Q: Why is "Using $F\Delta t$ with a single force value when the force varies (instead of the area under the F–t graph)" a common mistake in Momentum and Impulse?

Why it happens: The formula F t is memorised without the condition that F must be constant; students read the peak force off the graph and multiply by the total time. How to avoid it: For a **varying** force the impulse is the **area under the force–time graph**. For a triangular graph that area is \tfrac12 × base × height — half of (peak force × time), not the full rectangle. Count squares for a curved graph. Source: Edexcel IAL Physics Unit 4 examiner reports

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Short Notes - Momentum and Impulse

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Detailed Study Notes

Detailed notes on Further Mechanics, Fields and Particles for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Momentum and Impulse — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 4)

Impulse as $F\Delta t = \Delta p$ and as the area under a force–time graph; Newton's second law as the rate of change of momentum, $F = \Delta p/\Delta t$ (synoptic with Unit 1); conservation of linear momentum in two dimensions by resolving into components; elastic versus inelastic collisions and the kinetic-energy test; and kinetic energy from momentum, $E_k = p^2/2m$ . Includes Core Practicals 9 and 10.

At a glance

Impulse $= F\Delta t = \Delta p$ (change of momentum). Unit: $\text{N s} = \text{kg m s}^{-1}$ . Impulse and momentum are vectors — keep the signs.
For a varying force, impulse $=$ the AREA under the force–time graph — not $F\Delta t$ with a single force. A triangular graph gives $\tfrac12 \times \text{base} \times \text{height}$ .
Newton's second law in full is $F = \dfrac{\Delta p}{\Delta t}$ (rate of change of momentum). For constant mass this reduces to $\Sigma F = ma$ — a synoptic link back to Unit 1.
Momentum is conserved in every collision and explosion provided no external resultant force acts (a closed system).
In 2D, resolve momentum into components: the x- and y-momenta are conserved separately; the momentum vectors form a closed vector triangle.
Elastic collision: both momentum AND kinetic energy conserved. Inelastic collision: momentum conserved but kinetic energy is NOT (lost to heat/sound/deformation). Perfectly inelastic ⇒ objects stick together.
Test for elastic: compare total kinetic energy before and after. Momentum conservation alone never tells you the type.
Kinetic energy from momentum: $E_k = \dfrac{p^2}{2m}$ — derived from $E_k = \tfrac12 mv^2$ and $p = mv$ ; central to the particle-physics work later in Unit 4.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

4.x — Use the relationship $F\Delta t = \Delta p$ (impulse = change in momentum), and interpret the area under a force–time graph as impulse for a constant or varying force.
4.x — Express Newton's second law as $F = \dfrac{\Delta p}{\Delta t}$ , the rate of change of momentum, and relate it to $\Sigma F = ma$ for constant mass (synoptic with Unit 1).
4.x — Apply the principle of conservation of linear momentum to collisions and explosions in two dimensions by resolving momentum into perpendicular components.
4.x — Distinguish elastic from inelastic collisions and test which applies by comparing total kinetic energy before and after; identify perfectly inelastic collisions.
4.x — Use $E_k = \dfrac{p^2}{2m}$ to relate kinetic energy and momentum.
CP9 — Investigate the relationship between force and rate of change of momentum (force sensor + motion/light gates); plot $F$ against $\Delta p/\Delta t$ , gradient = 1.
CP10 — Analyse collisions using ICT / video capture; classify as elastic or inelastic and evaluate the uncertainty arising from the frame rate.

Impulse: $F\Delta t = \Delta p$ and the area under a force–time graph

Impulse is force × time and equals the change in momentum — but for a varying force it is the area under the F–t graph.

Momentum is $p = mv$ , a vector measured in $\text{kg m s}^{-1}$ . When a resultant force acts on an object, it changes the object's momentum. The impulse of the force is defined as the force multiplied by the time for which it acts:

$\text{impulse} = F\Delta t$

The central result of this topic is the impulse–momentum relationship: the impulse equals the change in momentum produced,

$F\Delta t = \Delta p = mv - mu.$

Impulse and change of momentum are the same vector quantity and share the unit $\text{N s}$ , which is identical to $\text{kg m s}^{-1}$ . Because they are vectors, you must choose a positive direction and keep the signs: when an object reverses, $\Delta v = v - (-u)$ , so the speeds add.

The crucial subtlety — a varying force. The formula $F\Delta t$ only works when the force is constant. In a real impact (a bat striking a ball, a car crumpling) the force rises and falls during the contact. For a varying force the impulse is the area under the force–time graph:

$\text{impulse} = \int F\,\mathrm{d}t = \text{area under the } F\text{–}t \text{ graph}.$

Think of the graph as thin vertical strips: each strip of width $\delta t$ has area $F\,\delta t$ (a small impulse), and summing all the strips gives the total area — the total impulse and hence $\Delta p$ . For a triangular graph the area is $\tfrac12 \times \text{base} \times \text{height}$ ; for a curved graph, count squares or use the trapezium rule.

The impulse delivered by a varying force is the area between the force–time line and the time axis. For this triangular pulse the area is ½ × base × height — one half of (peak force × contact time), NOT the full rectangle. That area equals the change in momentum, Δp.

Why impulse matters practically. Because $\Delta p$ is fixed by the mass and the change in velocity, and $F = \Delta p / \Delta t$ , spreading the same momentum change over a longer time gives a smaller average force. This is the physics of crumple zones, air bags, crash mats and bending your knees when you land: extend $\Delta t$ to cut the peak force.

$F\Delta t = \Delta p = mv - mu$ ; unit of impulse $= \text{N s} = \text{kg m s}^{-1}$ .
For a varying force, impulse = area under the F–t graph (triangle = $\tfrac12 bh$ ; curve = count squares).
Impulse and momentum are vectors — reversals make the speeds add.
Larger $\Delta t$ for the same $\Delta p$ ⇒ smaller force (crumple zones, air bags).

See the full worked example for momentum and impulse →

Newton's second law as $F = \Delta p / \Delta t$ (synoptic with Unit 1)

The full Newton's second law is the rate of change of momentum; $\Sigma F = ma$ is just the constant-mass version.

In Unit 1 you met Newton's second law as $\Sigma F = ma$ . Its full and more fundamental statement is in terms of momentum: the resultant force on an object equals the rate of change of its momentum.

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

This is synoptic — the same law, expressed in the language of momentum. Rearranging gives the impulse form $F\Delta t = \Delta p$ directly, so the two ideas are one.

Recovering $\Sigma F = ma$ . For an object of constant mass, $\Delta p = m\Delta v$ , so

$F = \frac{\Delta p}{\Delta t} = \frac{m\Delta v}{\Delta t} = m\frac{\Delta v}{\Delta t} = ma,$

since $\Delta v / \Delta t$ is the acceleration. So $\Sigma F = ma$ is simply the special case of the momentum law when the mass does not change. The momentum form is more general because it also handles situations where mass changes — for example a jet of water hitting a wall, or a rocket ejecting fuel, where you work with the rate of change of momentum ( $\dfrac{\Delta m}{\Delta t}\,v$ for a mass flow at speed $v$ ).

Core Practical 9 — force and rate of change of momentum. CP9 tests this law directly. A trolley is pulled by a measured force (a force sensor) while its motion is tracked with a motion sensor or light gates, giving the momentum at successive times and hence $\Delta p/\Delta t$ . Plotting the resultant force $F$ (y-axis) against $\Delta p/\Delta t$ (x-axis):

comparing $F = \dfrac{\Delta p}{\Delta t}$ with $y = mx$ , the graph is a straight line through the origin;
the gradient is 1 (dimensionless).

A gradient of 1 within experimental uncertainty confirms $F = \Delta p/\Delta t$ . Sources of uncertainty include friction (so the measured resultant force differs from the applied force) and the timing resolution of the sensors.

Newton's second law in full: $F = \dfrac{\Delta p}{\Delta t}$ (rate of change of momentum).
Constant mass ⇒ $\Delta p = m\Delta v$ ⇒ $F = ma$ ; so $\Sigma F = ma$ is the special case.
The momentum form also handles changing mass (water jets, rockets).
CP9: plot $F$ vs $\Delta p/\Delta t$ → straight line through origin, gradient 1.

See the full worked example for momentum and impulse →

Conservation of momentum in two dimensions

Momentum is a vector: conserve the x- and y-components separately, or close a vector triangle.

The principle of conservation of linear momentum states that, in a closed system (no external resultant force), the total momentum before an interaction equals the total momentum after it. It follows from Newton's third law: during a collision the two bodies push on each other with equal and opposite forces for the same time, so they exchange equal and opposite impulses — one gains exactly the momentum the other loses.

Because momentum is a vector, in two dimensions it is conserved independently along two perpendicular axes. The standard method:

Choose two perpendicular axes (often the original line of motion, x, and the perpendicular, y).
Resolve every object's momentum into components: $p_x = mv\cos\theta$ and $p_y = mv\sin\theta$ .
Write one conservation equation per axis: total $p_x$ before $=$ total $p_x$ after, and total $p_y$ before $=$ total $p_y$ after.
Solve the two simultaneous equations for the unknowns (speeds and/or angles). Keep the signs — components on opposite sides of an axis are opposite in sign.

In a 2D collision the total momentum vector is unchanged, so the initial momentum equals the tip-to-tail sum of the two final momenta — the vectors form a closed triangle. Equivalently, momentum is conserved separately along x and along y (here the two vertical components are equal and opposite, cancelling to zero).

Explosions and recoil are the reverse of a sticking collision: a stationary body separates into parts, so the total momentum stays zero and the fragment momenta are equal and opposite. For two fragments they recoil in opposite directions; a lighter fragment moves faster (equal momentum, smaller mass). The vector triangle is the neat graphical version: because momentum before equals momentum after, the momentum vectors always form a closed figure.

Closed system (no external resultant force): total momentum before = after.
2D: conserve $p_x$ and $p_y$ separately using $p_x = mv\cos\theta$ , $p_y = mv\sin\theta$ .
The momentum vectors form a closed vector triangle.
Explosion/recoil: total momentum stays zero; fragments' momenta are equal and opposite.

See the full worked example for momentum and impulse →

Elastic and inelastic collisions

Momentum is conserved in both; kinetic energy is conserved only in elastic collisions — test by comparing total KE.

Every collision conserves momentum (given a closed system). What distinguishes collisions is what happens to the kinetic energy.

Elastic collision: total momentum and total kinetic energy are both conserved. No kinetic energy is transferred to other forms. (Collisions between hard objects like snooker balls are nearly elastic; collisions between subatomic particles can be perfectly elastic.)
Inelastic collision: total momentum is conserved but total kinetic energy is not — some is transferred to heat (internal energy), sound and permanent deformation. Most everyday collisions are inelastic.
Perfectly inelastic collision: the objects stick together and move off with a common velocity. This loses the maximum kinetic energy consistent with momentum conservation.

How to test whether a collision is elastic. Momentum conservation cannot tell you — it holds for every collision. You must compare the total kinetic energy before and after:

$E_k = \tfrac12 m v^2 \quad\text{for each object; sum them.}$

If total $E_k$ (after) $=$ total $E_k$ (before), the collision is elastic. If total $E_k$ has decreased, it is inelastic, and the difference is the energy transferred to heat, sound and deformation. Note that KE uses $v^2$ , so a negative velocity sign disappears once squared — never keep the minus sign inside the square.

Both collisions conserve momentum. In an elastic collision the total kinetic energy is also conserved and the objects separate. In a perfectly inelastic collision the objects stick together and move as one; momentum is still conserved but kinetic energy is lost to heat, sound and deformation.

Kinetic energy from momentum. Because $p = mv$ , we can write kinetic energy in terms of momentum. Substituting $v = p/m$ into $E_k = \tfrac12 mv^2$ :

$E_k = \tfrac12 m\left(\frac{p}{m}\right)^2 = \frac{p^2}{2m}.$

The form $E_k = \dfrac{p^2}{2m}$ is invaluable when you know an object's momentum rather than its speed — which is the usual situation in the particle-physics part of Unit 4 (e.g. a nucleus recoiling with a known momentum). Remember to square $p$ and divide by $2m$ .

Elastic: momentum AND kinetic energy conserved. Inelastic: only momentum conserved.
Perfectly inelastic: objects stick together (maximum KE lost).
Test elastic by comparing total $E_k$ before and after — $E_k$ uses $v^2$ , so signs vanish.
$E_k = \dfrac{p^2}{2m}$ — kinetic energy from momentum; square $p$ , divide by $2m$ .

See the full worked example for momentum and impulse →

Quick recap

Impulse $= F\Delta t = \Delta p$ (change of momentum), unit $\text{N s} = \text{kg m s}^{-1}$ ; for a varying force it is the AREA under the force–time graph.
Newton's second law in full is $F = \Delta p/\Delta t$ ; for constant mass this gives $\Sigma F = ma$ (synoptic with Unit 1).
Momentum is conserved in any closed system; in 2D resolve into perpendicular components and conserve each axis (closed vector triangle).
Explosions and recoil: total momentum stays zero, so fragment momenta are equal and opposite; the lighter fragment moves faster.
Momentum is conserved in ALL collisions; kinetic energy is conserved ONLY in elastic collisions — test by comparing total KE before and after.
Perfectly inelastic collisions: objects stick together and lose the most KE (to heat, sound, deformation).
$E_k = \dfrac{p^2}{2m}$ — kinetic energy from momentum; square $p$ , divide by $2m$ . Quote answers to 3 s.f. with units.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Impulse $= F\Delta t = \Delta p = mv - mu$ ; unit $\text{N s} = \text{kg m s}^{-1}$ .
Varying force: impulse $=$ area under the force–time graph (triangle $= \tfrac12 bh$ ).
Newton's second law: $F = \dfrac{\Delta p}{\Delta t}$ ; constant mass ⇒ $\Sigma F = ma$ .
Conservation of momentum: total $p$ before $=$ total $p$ after (closed system); in 2D conserve $p_x$ and $p_y$ separately.
Elastic: momentum AND kinetic energy conserved. Inelastic: momentum conserved, KE not (→ heat/sound/deformation).
Perfectly inelastic ⇒ objects stick together and move with a common velocity.
$E_k = \dfrac{p^2}{2m}$ (from $E_k = \tfrac12 mv^2$ and $p = mv$ ).
CP9: plot $F$ vs $\Delta p/\Delta t$ → straight line through origin, gradient 1.

How it’s examined

Momentum and impulse sit on the Unit 4 paper (WPH14, Further Mechanics, Fields and Particles), an IA2 theory paper of structured questions with short calculations, 'show that' items, graph work and extended multi-step problems. Typical formats: define impulse or classify a collision (1–3 marks, AO1); a force-from-impulse calculation using $F = \Delta p/\Delta t$ , watching the sign on a reversal (3–4 marks, M and A marks); read an impulse as the AREA under a force–time graph (2–4 marks); a two-dimensional collision or explosion where you must resolve momentum into components and conserve each axis, or close a vector triangle (5–6 marks); an elastic-versus-inelastic test comparing total kinetic energy (4–6 marks); use of $E_k = p^2/2m$ in a particle-physics context (2–3 marks); and a practical/data item on CP9 (gradient of $F$ against $\Delta p/\Delta t$ is 1) or CP10 (video analysis, frame-rate uncertainty). This topic is SYNOPTIC with Unit 1: Newton's second law is genuinely $F = \Delta p/\Delta t$ , and momentum conservation follows from Newton's third law. ALWAYS show the equation, substitution with units, and a 3 s.f. answer; keep vector signs; for 'show that' work to an extra significant figure.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 4 (WPH14) examiner reports; Edexcel IAL Physics Unit 4 sample assessment materials and mark schemes. Last reviewed 2026-07-05.

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Step-by-step worked examples — Momentum and Impulse

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

Question type:

1Impulse from force and contact time (2 marks)
Getting startedDirect calculation• impulse, AO2, Unit 4
Question
A constant force of $250\ \text{N}$ acts on a trolley for $0.040\ \text{s}$ . Calculate the impulse delivered to the trolley. (2 marks)
Step-by-step solution
1. Step 1
  For a constant force, impulse is force multiplied by the time for which it acts.
  $\text{impulse} = F\Delta t$
2. Step 2
  Substitute $F = 250\ \text{N}$ and $\Delta t = 0.040\ \text{s}$ and evaluate to 3 significant figures.
  $F\Delta t = 250 \times 0.040 = 10.0\ \text{N s}$
Answer
Impulse $= 10.0\ \text{N s}$ (equivalently $10.0\ \text{kg m s}^{-1}$ ), in the direction of the force.
Examiner tip
Mark scheme: (1) use of $\text{impulse} = F\Delta t$ ; (2) $10.0\ \text{N s}$ to 3 s.f. with a correct unit. Impulse and change of momentum share the same unit — $\text{N s}$ and $\text{kg m s}^{-1}$ are identical.
2Kinetic energy from momentum, $E_k = p^2/2m$ (2 marks)
Getting startedDirect calculation• kinetic energy, momentum, AO2, Unit 4
Question
A proton of mass $1.67 \times 10^{-27}\ \text{kg}$ has a momentum of $2.00 \times 10^{-21}\ \text{kg m s}^{-1}$ . Calculate its kinetic energy. (2 marks)
Step-by-step solution
1. Step 1
  Kinetic energy can be written in terms of momentum using $E_k = \dfrac{p^2}{2m}$ (from $E_k = \tfrac{1}{2}mv^2$ and $p = mv$ ).
  $E_k = \frac{p^2}{2m}$
2. Step 2
  Substitute $p = 2.00 \times 10^{-21}\ \text{kg m s}^{-1}$ and $m = 1.67 \times 10^{-27}\ \text{kg}$ .
  $E_k = \frac{(2.00 \times 10^{-21})^2}{2 \times 1.67 \times 10^{-27}} = \frac{4.00 \times 10^{-42}}{3.34 \times 10^{-27}} = 1.20 \times 10^{-15}\ \text{J}$
Answer
$E_k = 1.20 \times 10^{-15}\ \text{J}$ .
Examiner tip
Mark scheme: (1) use of $E_k = p^2/2m$ ; (2) $1.20 \times 10^{-15}\ \text{J}$ to 3 s.f. Squaring $p$ is essential — a common slip is to divide $p$ (not $p^2$ ) by $2m$ . This form appears throughout the Unit 4 particle-physics section.
3Elastic or inelastic? (3 marks)
Getting startedIdentify & classify• elastic, inelastic, AO1, Unit 4
Question
In a collision the total momentum before equals the total momentum after, but the total kinetic energy after is less than before. State, with a reason, whether the collision is elastic or inelastic, and say what the two conservation rules are for such a collision. (3 marks)
Step-by-step solution
1. Step 1
  Momentum is conserved in every collision (provided no external resultant force acts), so that fact alone does not classify the collision.
2. Step 2
  A collision is elastic only if total kinetic energy is also conserved. Here the kinetic energy has fallen, so kinetic energy is not conserved.
3. Step 3
  Therefore the collision is inelastic: momentum is conserved but kinetic energy is not — the 'missing' kinetic energy is transferred to heat, sound and deformation.
Answer
The collision is inelastic: momentum is conserved (as in all collisions) but total kinetic energy is not conserved — some is transferred to heat, sound and deformation.
Examiner tip
Mark scheme: (1) momentum conserved in all collisions; (2) KE decreased ⇒ KE not conserved; (3) conclusion 'inelastic'. The discriminator examiners look for is that momentum conservation does NOT distinguish elastic from inelastic — only the kinetic-energy test does.
4Force from a change of momentum (4 marks)
Building confidenceMulti-step problem• impulse, Newton's second law, AO2, Unit 4
Question
A $0.058\ \text{kg}$ tennis ball travelling at $9.0\ \text{m s}^{-1}$ is struck and returns straight back along its path at $22.0\ \text{m s}^{-1}$ . The racket is in contact with the ball for $5.0 \times 10^{-3}\ \text{s}$ . Calculate the average force the racket exerts on the ball. (4 marks)
Step-by-step solution
1. Step 1
  Take the return direction (after the strike) as positive. The initial velocity is then $-9.0\ \text{m s}^{-1}$ and the final velocity $+22.0\ \text{m s}^{-1}$ , because the ball reverses direction.
2. Step 2
  Find the change in momentum, $\Delta p = mv - mu$ , keeping the signs.
  $\Delta p = 0.058 \times (22.0 - (-9.0)) = 0.058 \times 31.0 = 1.798\ \text{kg m s}^{-1}$
3. Step 3
  Newton's second law in impulse form gives $F\Delta t = \Delta p$ . Rearrange for the average force.
  $F = \frac{\Delta p}{\Delta t}$
4. Step 4
  Substitute and evaluate to 3 significant figures.
  $F = \frac{1.798}{5.0 \times 10^{-3}} = 360\ \text{N}$
Answer
$F = 360\ \text{N}$ (to 3 s.f.), acting in the direction the ball leaves the racket.
Examiner tip
Mark scheme: (1) assign a positive direction / reverse one velocity's sign; (2) $\Delta p = 1.80\ \text{kg m s}^{-1}$ using $22 - (-9) = 31$ ; (3) use of $F = \Delta p/\Delta t$ ; (4) $360\ \text{N}$ . Using $22 - 9 = 13$ is the fatal slip — because the ball reverses, the speeds ADD.
5Impulse as the area under a force–time graph (4 marks)
Building confidenceGraph or diagram• impulse, force-time graph, AO2, Unit 4
Question
A ball of mass $0.42\ \text{kg}$ is initially at rest. A varying force acts on it for $0.30\ \text{s}$ : the force rises linearly from $0$ to a peak of $120\ \text{N}$ at $0.20\ \text{s}$ , then falls linearly back to $0$ at $0.30\ \text{s}$ (a triangular force–time graph). Calculate (a) the impulse delivered and (b) the final speed of the ball. (4 marks)
Step-by-step solution
1. Step 1
  For a varying force the impulse is the area under the force–time graph, not $F\Delta t$ with a single force value. The graph is a triangle, so use $\tfrac{1}{2} \times \text{base} \times \text{height}$ .
  $\text{impulse} = \text{area} = \tfrac{1}{2}\,(\Delta t)(F_{peak})$
2. Step 2
  The base is the total time $0.30\ \text{s}$ and the height is the peak force $120\ \text{N}$ .
  $\text{impulse} = \tfrac{1}{2} \times 0.30 \times 120 = 18.0\ \text{N s}$
3. Step 3
  (b) Impulse equals the change in momentum. The ball starts from rest, so $\Delta p = mv - 0 = mv$ .
  $\text{impulse} = \Delta p = mv$
4. Step 4
  Rearrange for the final speed $v$ and evaluate.
  $v = \frac{\text{impulse}}{m} = \frac{18.0}{0.42} = 42.9\ \text{m s}^{-1}$
Answer
(a) Impulse $= 18.0\ \text{N s}$ . (b) $v = 42.9\ \text{m s}^{-1}$ in the direction of the force.
Examiner tip
Mark scheme: (a) 2 marks — recognise impulse = area under the F–t graph; area of triangle $= \tfrac12 \times 0.30 \times 120 = 18.0\ \text{N s}$ . (b) 2 marks — impulse $= \Delta p = mv$ ; $v = 42.9\ \text{m s}^{-1}$ . The classic error is $F\Delta t = 120 \times 0.30$ , which double-counts because the force is not constant — the area (a triangle) is half of that.
6Recoil in an explosion using conservation of momentum (5 marks)
Building confidenceMulti-step problem• conservation of momentum, explosion, kinetic energy, AO2, Unit 4
Question
A stationary nucleus of mass $3.8 \times 10^{-25}\ \text{kg}$ decays, emitting an alpha particle of mass $6.6 \times 10^{-27}\ \text{kg}$ at a speed of $1.5 \times 10^{7}\ \text{m s}^{-1}$ . (a) Calculate the recoil speed of the remaining nucleus. (b) Calculate the kinetic energy of the recoiling nucleus using $E_k = p^2/2m$ . (5 marks)
Step-by-step solution
1. Step 1
  (a) Before the decay everything is at rest, so the total momentum is zero. Momentum is conserved (internal forces only), so the total after the decay is also zero. The recoiling nucleus has mass $m_N = 3.8 \times 10^{-25} - 6.6 \times 10^{-27} = 3.734 \times 10^{-25}\ \text{kg}$ .
  $0 = m_\alpha v_\alpha + m_N v_N$
2. Step 2
  The alpha particle's momentum equals the recoiling nucleus's momentum in magnitude (opposite sign). Find that momentum.
  $p_\alpha = (6.6 \times 10^{-27})(1.5 \times 10^{7}) = 9.9 \times 10^{-20}\ \text{kg m s}^{-1}$
3. Step 3
  The nucleus carries $-9.9 \times 10^{-20}\ \text{kg m s}^{-1}$ . Its recoil speed is that momentum divided by its mass.
  $v_N = \frac{9.9 \times 10^{-20}}{3.734 \times 10^{-25}} = 2.65 \times 10^{5}\ \text{m s}^{-1}$
4. Step 4
  (b) The nucleus and the alpha particle have equal-magnitude momenta, so $p_N = 9.9 \times 10^{-20}\ \text{kg m s}^{-1}$ . Use $E_k = p^2/2m$ with the nucleus's mass.
  $E_k = \frac{p_N^2}{2m_N}$
5. Step 5
  Substitute and evaluate to 3 significant figures.
  $E_k = \frac{(9.9 \times 10^{-20})^2}{2 \times 3.734 \times 10^{-25}} = 1.31 \times 10^{-14}\ \text{J}$
Answer
(a) $v_N = 2.65 \times 10^{5}\ \text{m s}^{-1}$ in the opposite direction to the alpha particle. (b) $E_k = 1.31 \times 10^{-14}\ \text{J}$ .
Examiner tip
Mark scheme: (a) 3 marks — total momentum zero; $p_\alpha = 9.9 \times 10^{-20}\ \text{kg m s}^{-1}$ ; recoil speed $= 2.65 \times 10^{5}\ \text{m s}^{-1}$ using the correct residual mass. (b) 2 marks — use of $E_k = p^2/2m$ ; $1.31 \times 10^{-14}\ \text{J}$ . Using the original nucleus mass ( $3.8 \times 10^{-25}$ ) instead of the residual mass is the common error.
7Two-dimensional collision — resolving momentum (6 marks)
StretchMulti-step problem• conservation of momentum, 2D, vectors, AO2, Unit 4
Question
A ball A of mass $0.20\ \text{kg}$ moving east at $4.0\ \text{m s}^{-1}$ strikes a stationary ball B of mass $0.20\ \text{kg}$ . After the collision A moves off at $30°$ north of east and B at $60°$ south of east. Calculate the speeds of A and B after the collision. (6 marks)
Step-by-step solution
1. Step 1
  Momentum is conserved separately along the east (x) and north (y) axes. Total momentum before is entirely along x: $p_x = (0.20)(4.0) = 0.80\ \text{kg m s}^{-1}$ ; $p_y = 0$ .
  $p_x = 0.80\ \text{kg m s}^{-1},\quad p_y = 0$
2. Step 2
  Let $v_A$ and $v_B$ be the speeds after the collision. Resolve the y-components: A goes $30°$ north of east ( $+v_A\sin 30°$ ), B goes $60°$ south of east ( $-v_B\sin 60°$ ). The y-momentum must total zero.
  $0.20\,v_A\sin 30° - 0.20\,v_B\sin 60° = 0$
3. Step 3
  The masses cancel: $v_A\sin 30° = v_B\sin 60°$ . So $v_A(0.500) = v_B(0.866)$ , giving $v_A = 1.732\,v_B$ .
  $v_A = \frac{\sin 60°}{\sin 30°}\,v_B = 1.732\,v_B$
4. Step 4
  Now resolve the x-components: A contributes $v_A\cos 30°$ , B contributes $v_B\cos 60°$ . The x-momentum must total $4.0\ \text{m s}^{-1}$ worth (dividing the earlier equation by the common mass $0.20$ ).
  $v_A\cos 30° + v_B\cos 60° = 4.0$
5. Step 5
  Substitute $v_A = 1.732\,v_B$ and $\cos 30° = 0.866$ , $\cos 60° = 0.500$ .
  $(1.732\,v_B)(0.866) + v_B(0.500) = 4.0 \;\Rightarrow\; 1.500\,v_B + 0.500\,v_B = 2.000\,v_B = 4.0$
6. Step 6
  Solve for $v_B$ , then for $v_A$ .
  $v_B = 2.00\ \text{m s}^{-1},\quad v_A = 1.732 \times 2.00 = 3.46\ \text{m s}^{-1}$
Answer
$v_A = 3.46\ \text{m s}^{-1}$ (at $30°$ north of east) and $v_B = 2.00\ \text{m s}^{-1}$ (at $60°$ south of east).
Examiner tip
Mark scheme: (1) resolve momentum into x and y and state each is conserved; (2) y-equation with opposite signs for the two balls; (3) combine to relate $v_A$ and $v_B$ ; (4) x-equation; (5)/(6) solve for $v_B = 2.00$ and $v_A = 3.46\ \text{m s}^{-1}$ . The killer error is treating the momenta as scalars — the y-components must carry opposite signs so they cancel to zero.
8Momentum vector triangle for an explosion (6 marks)
StretchGraph or diagram• conservation of momentum, 2D, vector triangle, AO2, Unit 4
Question
A stationary object explodes into two fragments. Fragment X of mass $0.30\ \text{kg}$ moves due north at $8.0\ \text{m s}^{-1}$ . Fragment Y of mass $0.40\ \text{kg}$ moves off in some direction. Using a momentum vector triangle, calculate the speed and direction of fragment Y. (6 marks)
Step-by-step solution
1. Step 1
  Before the explosion the total momentum is zero. Momentum is conserved, so the two fragment momenta must be equal and opposite — they add tip-to-tail to give zero, forming a closed triangle (here a straight line).
  $\vec{p}_X + \vec{p}_Y = 0$
2. Step 2
  Find the momentum of fragment X (due north).
  $p_X = (0.30)(8.0) = 2.4\ \text{kg m s}^{-1}\ \text{(north)}$
3. Step 3
  For the total to be zero, fragment Y must carry an equal-magnitude momentum in the opposite direction — due south.
  $p_Y = 2.4\ \text{kg m s}^{-1}\ \text{(south)}$
4. Step 4
  Find Y's speed from its momentum and mass.
  $v_Y = \frac{p_Y}{m_Y} = \frac{2.4}{0.40} = 6.0\ \text{m s}^{-1}$
5. Step 5
  State the direction: because X went north, Y must recoil due south (the two momentum vectors are anti-parallel and equal in size).
Answer
Fragment Y moves at $6.0\ \text{m s}^{-1}$ due south — directly opposite to fragment X, so the momentum vectors are equal and opposite and sum to zero.
Examiner tip
Mark scheme: (1) total momentum zero ⇒ momenta equal and opposite (closed vector triangle); (2) $p_X = 2.4\ \text{kg m s}^{-1}$ ; (3) $p_Y = 2.4\ \text{kg m s}^{-1}$ opposite direction; (4) $v_Y = 6.0\ \text{m s}^{-1}$ ; (5) direction due south. Note the fragments have different speeds ( $8.0$ vs $6.0$ ) but equal momenta — the lighter fragment moves faster.
9Core Practical 9 — force equals rate of change of momentum (6 marks)
StretchShow that / prove• Adapted from Edexcel IAL Physics Unit 4 (CP9)• Core Practical 9, Newton's second law, graph, AO3, Unit 4
Question
In Core Practical 9 a student uses a force sensor and a motion sensor to investigate the relationship between the resultant force $F$ on a trolley and its rate of change of momentum $\Delta p/\Delta t$ . They plot $F$ (y-axis) against $\Delta p/\Delta t$ (x-axis) and obtain a straight line through the origin. (a) State what gradient the theory predicts and why. (b) One data point is $F = 2.40\ \text{N}$ with $\Delta p/\Delta t = 2.35\ \text{N}$ . Comment on whether this supports the theory, and give one source of uncertainty. (6 marks)
Step-by-step solution
1. Step 1
  (a) Newton's second law in its full form states that the resultant force equals the rate of change of momentum: $F = \dfrac{\Delta p}{\Delta t}$ .
  $F = \frac{\Delta p}{\Delta t}$
2. Step 2
  Comparing with $y = mx$ , plotting $F$ against $\Delta p/\Delta t$ should give a straight line through the origin of gradient 1 (dimensionless). A gradient of 1 confirms $F = \Delta p/\Delta t$ .
  $\text{gradient} = 1$
3. Step 3
  (b) Compare the two values: $F = 2.40\ \text{N}$ and $\Delta p/\Delta t = 2.35\ \text{N}$ . Their ratio is $2.40/2.35 = 1.02$ , very close to the predicted gradient of 1.
  $\frac{F}{\Delta p/\Delta t} = \frac{2.40}{2.35} = 1.02$
4. Step 4
  The small difference (about 2%) is within experimental uncertainty, so the point supports the theory $F = \Delta p/\Delta t$ .
5. Step 5
  A source of uncertainty: friction in the trolley or pulley means the measured resultant force differs slightly from the applied force; or the timing resolution of the motion sensor limits the precision of $\Delta p/\Delta t$ .
Answer
(a) The predicted gradient is 1 (dimensionless), because Newton's second law gives $F = \Delta p/\Delta t$ . (b) The ratio $2.40/2.35 = 1.02 \approx 1$ , within ~2%, so the data support the theory; friction (or the timing resolution of the sensors) is a source of uncertainty.
Examiner tip
Mark scheme: (a) 3 marks — $F = \Delta p/\Delta t$ ; line through origin; gradient 1 (dimensionless). (b) 3 marks — ratio ≈ 1.02; conclusion 'supports theory within uncertainty'; one valid uncertainty (friction, sensor timing/resolution, alignment). This is a synoptic link to Unit 1: Newton's second law is genuinely $F = \Delta p/\Delta t$ , and $\Sigma F = ma$ is the constant-mass special case.

Model Answers — Momentum and Impulse

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

Question 1
2 marks
Q (2 marks). State what is meant by the impulse of a force and give its relationship to momentum. Include the unit of impulse.
Model answer
The impulse of a force is the product of the force and the time for which it acts, $F\Delta t$ . It is a vector quantity.

Impulse is equal to the change in momentum it produces: $F\Delta t = \Delta p = mv - mu$ The unit of impulse is the newton second ( $\text{N s}$ ), which is identical to $\text{kg m s}^{-1}$ .
Why this scores
Why this scores 2/2. (1) impulse = force × time ( $F\Delta t$ ); (2) impulse = change in momentum, with the correct unit $\text{N s}$ . Stating that impulse equals $\Delta p$ is the key link — many candidates define impulse but never connect it to momentum.
Question 2
3 marks
Q (3 marks). Distinguish between an elastic and an inelastic collision. State which quantity is conserved in both types, and what happens to the kinetic energy in an inelastic collision.
Model answer
In an elastic collision both total momentum and total kinetic energy are conserved.

In an inelastic collision total momentum is still conserved, but total kinetic energy is not conserved.

Linear momentum is conserved in both types (provided no external resultant force acts). In an inelastic collision the 'lost' kinetic energy is transferred to other forms — mainly heat (internal energy), sound and permanent deformation. Total energy is still conserved overall; only kinetic energy is reduced.
Why this scores
Why this scores 3/3. (1) elastic = both momentum and KE conserved; (2) inelastic = momentum conserved but KE not; (3) momentum conserved in both, and KE → heat/sound/deformation. The single most credited point is that momentum is conserved in every collision — it is KE that separates the two.
Question 3
3 marks
Q (3 marks). Show that the kinetic energy of an object of mass $m$ and momentum $p$ can be written as $E_k = \dfrac{p^2}{2m}$ .
Model answer
Start from the standard expression for kinetic energy: $E_k = \tfrac{1}{2}mv^2$

Momentum is defined as $p = mv$ , so the velocity is $v = \dfrac{p}{m}$ .

Substitute this into the kinetic-energy expression: $E_k = \tfrac{1}{2}m\left(\frac{p}{m}\right)^2 = \tfrac{1}{2}m \cdot \frac{p^2}{m^2} = \frac{p^2}{2m}$

Hence $E_k = \dfrac{p^2}{2m}$ , as required.
Why this scores
Why this scores 3/3. (1) start from $E_k = \tfrac12 mv^2$ ; (2) substitute $v = p/m$ ; (3) simplify correctly to $p^2/2m$ . A 'show that' must display every algebraic step — jumping straight to the result earns nothing.
Question 4
4 marks
Q (4 marks). A car of mass $1100\ \text{kg}$ travelling at $18\ \text{m s}^{-1}$ is stopped in a crash lasting $0.15\ \text{s}$ . (a) Calculate the magnitude of the average force on the car. (b) Explain, in terms of impulse, how a longer crumple time reduces the force on the occupants.
Model answer
(a) Change in momentum: $\Delta p = mv - mu = 1100 \times (0 - 18) = -1.98 \times 10^{4}\ \text{kg m s}^{-1}$ Average force from $F = \dfrac{\Delta p}{\Delta t}$ : $F = \frac{-1.98 \times 10^{4}}{0.15} = -1.32 \times 10^{5}\ \text{N}$ The magnitude of the average force is $1.32 \times 10^{5}\ \text{N}$ , directed opposite to the motion.

(b) The change in momentum $\Delta p$ is fixed by the mass and the speed change — the occupants must lose the same momentum however they stop. Since $F = \Delta p/\Delta t$ , increasing the time $\Delta t$ over which that momentum change occurs decreases the average force $F$ . A crumple zone extends the collision time, so the force on the occupants, and the risk of injury, is reduced.
Why this scores
Why this scores 4/4. (a) 2 marks — $\Delta p = -1.98 \times 10^{4}\ \text{kg m s}^{-1}$ ; $F = 1.32 \times 10^{5}\ \text{N}$ . (b) 2 marks — $\Delta p$ fixed; larger $\Delta t$ ⇒ smaller $F$ . The high-value point is that $\Delta p$ is unchanged, so the benefit comes purely from extending the time.
Question 5
4 marks
Q (4 marks). A golf club strikes a ball and the force on the ball varies with time as a smooth curve, rising to a peak and falling back to zero. (a) Explain why the impulse on the ball equals the area under the force–time graph. (b) State how you would obtain the impulse from a graph whose shape is not a simple triangle.
Model answer
(a) Over a very short time interval $\delta t$ the force is approximately constant, so the small impulse is $F\,\delta t$ — the area of a thin strip under the graph. Summing all such strips over the whole contact time adds up all the small impulses, which is the total area under the force–time graph. Because impulse equals the change in momentum, the area also gives $\Delta p$ : $\text{impulse} = \sum F\,\delta t = \text{area under } F\text{–}t\text{ graph} = \Delta p$

(b) For a curved (non-triangular) graph, find the area by counting squares under the curve (multiplying the number of squares by the value of one square in $\text{N s}$ ), or by dividing the area into strips and using the trapezium rule. This handles any shape, unlike $F\Delta t$ , which only works for a constant force.
Why this scores
Why this scores 4/4. (a) 2 marks — sum of $F\,\delta t$ strips = area; area = impulse = $\Delta p$ . (b) 2 marks — count squares / trapezium rule. The examiner wants recognition that $F\Delta t$ fails for a varying force and the area is the correct general method.
Question 6
4 marks
Q (4 marks). A moving snooker ball strikes an identical stationary ball off-centre; the two balls move away at different angles. Explain how the principle of conservation of momentum is applied to this two-dimensional collision.
Model answer
Momentum is a vector, so it is conserved independently in two perpendicular directions. Choose two axes — for example, the original direction of motion (x) and the direction perpendicular to it (y).

Along the x-axis: the total x-component of momentum after the collision equals the total x-component before (which is just the momentum of the incoming ball).

Along the y-axis: the total y-component of momentum before is zero, so the y-components of the two balls after the collision must be equal and opposite — they cancel.

By resolving each ball's momentum into $p\cos\theta$ (along x) and $p\sin\theta$ (along y) and writing one conservation equation per axis, the two unknown speeds (or a speed and an angle) can be found. Equivalently, the momentum vectors form a closed vector triangle, since the total momentum is unchanged.
Why this scores
Why this scores 4/4. (1) momentum is a vector; (2) conserved separately in x and y; (3) resolve each ball into components ( $p\cos\theta$ , $p\sin\theta$ ); (4) y-components cancel / vector triangle closes. Treating the problem as one-dimensional is the level-capping error.
Question 7
5 marks
Q (5 marks). (a) State Newton's second law of motion in terms of momentum. (b) Show that, for an object of constant mass, this leads to $F = ma$ . (c) A jet of water leaves a hose horizontally at $15\ \text{m s}^{-1}$ , delivering $8.0\ \text{kg}$ of water each second onto a wall, where it stops. Calculate the average horizontal force on the wall.
Model answer
(a) The resultant force on an object is equal to the rate of change of its momentum: $F = \frac{\Delta p}{\Delta t}$

(b) For constant mass, $\Delta p = m\Delta v$ , so $F = \frac{m\Delta v}{\Delta t} = m\frac{\Delta v}{\Delta t} = ma$ since $\dfrac{\Delta v}{\Delta t}$ is the acceleration $a$ .

(c) Each second, $8.0\ \text{kg}$ of water is brought from $15\ \text{m s}^{-1}$ to rest, so the momentum destroyed per second is $\frac{\Delta p}{\Delta t} = \frac{(8.0)(15)}{1.0} = 120\ \text{kg m s}^{-1}\ \text{per second}$ Therefore the average force on the water is $120\ \text{N}$ , and by Newton's third law the water exerts an equal and opposite force of $120\ \text{N}$ on the wall.
Why this scores
Why this scores 5/5. (a) 1 mark — $F = \Delta p/\Delta t$ . (b) 2 marks — substitute $\Delta p = m\Delta v$ and identify $\Delta v/\Delta t = a$ . (c) 2 marks — rate of change of momentum $= 120\ \text{kg m s}^{-1}\,\text{s}^{-1}$ ; force $= 120\ \text{N}$ . This part (c) is the synoptic 'rate of mass flow × velocity' application of the momentum form.
Question 8
6 marks
Q (6 marks). A $0.16\ \text{kg}$ puck moving at $5.0\ \text{m s}^{-1}$ strikes a stationary $0.16\ \text{kg}$ puck head-on. After the collision the first puck is stationary and the second moves at $5.0\ \text{m s}^{-1}$ . (a) Show that momentum is conserved. (b) Determine, by calculation, whether the collision is elastic.
Model answer
(a) Take the initial direction as positive. Momentum before (only the first puck moves): $p_{before} = (0.16)(5.0) + (0.16)(0) = 0.80\ \text{kg m s}^{-1}$ Momentum after (only the second puck moves): $p_{after} = (0.16)(0) + (0.16)(5.0) = 0.80\ \text{kg m s}^{-1}$ Since $p_{before} = p_{after} = 0.80\ \text{kg m s}^{-1}$ , momentum is conserved.

(b) A collision is elastic only if total kinetic energy is also conserved. KE before: $E_{k,before} = \tfrac{1}{2}(0.16)(5.0)^2 = 2.0\ \text{J}$ KE after: $E_{k,after} = \tfrac{1}{2}(0.16)(5.0)^2 = 2.0\ \text{J}$ Since the total kinetic energy is unchanged ( $2.0\ \text{J} = 2.0\ \text{J}$ ) as well as the momentum, the collision is elastic.
Why this scores
Why this scores 6/6. (a) 3 marks — momentum before $= 0.80$ , after $= 0.80\ \text{kg m s}^{-1}$ , conclusion 'conserved'. (b) 3 marks — KE before $= 2.0\ \text{J}$ , KE after $= 2.0\ \text{J}$ , conclusion 'elastic because KE conserved'. This equal-mass head-on swap is the classic perfectly-elastic result: the velocities are exchanged.
Question 9
6 marks
Q (6 marks). A ball A of mass $0.50\ \text{kg}$ moving at $6.0\ \text{m s}^{-1}$ collides with a stationary ball B of mass $0.50\ \text{kg}$ . After the collision A moves off at $37°$ to its original direction and B moves off at $53°$ on the other side. Calculate the speed of ball A after the collision. (Take $\sin 37° = 0.60$ , $\cos 37° = 0.80$ , $\sin 53° = 0.80$ , $\cos 53° = 0.60$ .)
Model answer
Momentum is conserved separately along the original direction (x) and perpendicular to it (y). Let $v_A$ and $v_B$ be the speeds after the collision. The masses ( $0.50\ \text{kg}$ ) are equal and cancel throughout, so I work with velocities.

Perpendicular (y) direction — total before is zero: $v_A\sin 37° - v_B\sin 53° = 0 \;\Rightarrow\; 0.60\,v_A = 0.80\,v_B \;\Rightarrow\; v_B = 0.75\,v_A$

Original (x) direction — total before is $6.0\ \text{m s}^{-1}$ worth of velocity: $v_A\cos 37° + v_B\cos 53° = 6.0 \;\Rightarrow\; 0.80\,v_A + 0.60\,v_B = 6.0$

Substitute $v_B = 0.75\,v_A$ : $0.80\,v_A + 0.60(0.75\,v_A) = 0.80\,v_A + 0.45\,v_A = 1.25\,v_A = 6.0$ $v_A = \frac{6.0}{1.25} = 4.8\ \text{m s}^{-1}$

So ball A moves off at $4.8\ \text{m s}^{-1}$ (and $v_B = 0.75 \times 4.8 = 3.6\ \text{m s}^{-1}$ ).
Why this scores
Why this scores 6/6. (1) state momentum conserved in x and y; (2) perpendicular equation with opposite signs; (3) relate $v_B$ to $v_A$ ; (4) original-direction equation; (5) substitute; (6) $v_A = 4.8\ \text{m s}^{-1}$ . Note $37° + 53° = 90°$ : this equal-mass collision has the two balls departing at right angles, the signature of an elastic collision between equal masses.
Question 10
6 marks
Q (6 marks). In Core Practical 10 a student films a collision between two trolleys and uses video analysis to measure their velocities before and after. Explain how the student would (a) test whether momentum is conserved, (b) classify the collision as elastic or inelastic, and (c) reduce the uncertainty in the measured velocities.
Model answer
(a) Testing momentum conservation. From the video, measure the mass of each trolley and read off the velocity of each before and after the collision (using the change in position between frames and the known frame rate for time). Calculate the total momentum before ( $m_1u_1 + m_2u_2$ ) and the total momentum after ( $m_1v_1 + m_2v_2$ ), keeping the direction signs. If the two totals are equal (within experimental uncertainty), momentum is conserved.

(b) Classifying the collision. Calculate the total kinetic energy before and after using $E_k = \tfrac12 mv^2$ for each trolley. If the total kinetic energy is (within uncertainty) unchanged, the collision is elastic; if the kinetic energy has decreased, it is inelastic — momentum is conserved in either case.

(c) Reducing uncertainty. Use a higher frame rate so each velocity is found over more, finer time intervals, reducing the timing uncertainty; include a clear scale marker in shot and film perpendicular to the motion to avoid parallax; and repeat the measurement and average. The frame rate sets the time resolution, so a faster camera gives smaller uncertainty in $\Delta t$ and hence in velocity.
Why this scores
Why this reaches the top band (5–6/6). (a) measure masses and before/after velocities from frames; compare total momentum before and after. (b) compare total KE before and after ⇒ elastic if unchanged, inelastic if reduced. (c) higher frame rate (time resolution), scale marker/perpendicular filming (parallax), repeats. The discriminator is linking the frame rate to the timing uncertainty in the velocity, and stressing momentum is conserved for both collision types.

Key Formulae — Momentum and Impulse

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

Impulse = change in momentum
$F\Delta t = \Delta p = mv - mu$
$F$
average resultant force / $\text{N}$
$\Delta t$
time for which the force acts / $\text{s}$
$\Delta p$
change in momentum / $\text{kg m s}^{-1}$ (or $\text{N s}$ )
$u, v$
initial and final velocity / $\text{m s}^{-1}$
When to use
When a force acts for a measured time — to find the force, the impulse, or the velocity change. Keep velocity signs; a reversal makes the speeds add. Valid for a constant force; for a varying force use the area under the force–time graph instead.
Example
A racket delivering $\Delta p = 1.80\ \text{kg m s}^{-1}$ in $5.0 \times 10^{-3}\ \text{s}$ exerts $F = 1.80/(5.0 \times 10^{-3}) = 360\ \text{N}$ .
Impulse as area under a force–time graph
$\text{impulse} = \int F \, \mathrm{d}t = \text{area under the } F\text{–}t \text{ graph}$
$F$
instantaneous force / $\text{N}$
$t$
time / $\text{s}$
$area$
region between the $F$ – $t$ line and the time axis / $\text{N s}$
When to use
When the force is NOT constant (a curved or triangular force–time graph). Find the area geometrically (triangle $= \tfrac12 bh$ , trapezium, or by counting squares) — this equals the impulse and hence $\Delta p$ .
Example
A triangular F–t graph peaking at $120\ \text{N}$ over $0.30\ \text{s}$ gives impulse $= \tfrac12 \times 0.30 \times 120 = 18.0\ \text{N s}$ .
Newton's second law (rate of change of momentum)
$F = \frac{\Delta p}{\Delta t}$
$F$
resultant force / $\text{N}$
$\Delta p$
change in momentum / $\text{kg m s}^{-1}$
$\Delta t$
time interval / $\text{s}$
When to use
The full statement of Newton's second law (synoptic with Unit 1). Use it whenever mass may change, or when working directly with momentum; for constant mass it reduces to $\Sigma F = ma$ . Core Practical 9 verifies it: a plot of $F$ against $\Delta p/\Delta t$ has gradient 1.
Example
A trolley whose momentum changes by $6.0\ \text{kg m s}^{-1}$ in $2.0\ \text{s}$ experiences a resultant force $F = 6.0/2.0 = 3.0\ \text{N}$ .
Linear momentum
$p = mv$
$p$
momentum (a vector) / $\text{kg m s}^{-1}$
$m$
mass / $\text{kg}$
$v$
velocity / $\text{m s}^{-1}$
When to use
To calculate momentum and to set up conservation-of-momentum equations (total before = total after). In two dimensions, resolve into components: $p_x = mv\cos\theta$ , $p_y = mv\sin\theta$ , and conserve each axis separately.
Example
A $0.20\ \text{kg}$ ball at $4.0\ \text{m s}^{-1}$ has $p = 0.20 \times 4.0 = 0.80\ \text{kg m s}^{-1}$ .
Kinetic energy in terms of momentum
$E_k = \frac{p^2}{2m}$
$E_k$
kinetic energy / $\text{J}$
$p$
momentum / $\text{kg m s}^{-1}$
$m$
mass / $\text{kg}$
When to use
When you know an object's momentum (rather than its velocity) and need its kinetic energy — very common in particle physics. Derived from $E_k = \tfrac12 mv^2$ and $p = mv$ : substituting $v = p/m$ gives $E_k = p^2/2m$ . Remember to square $p$ .
Example
A proton with $p = 2.00 \times 10^{-21}\ \text{kg m s}^{-1}$ and $m = 1.67 \times 10^{-27}\ \text{kg}$ has $E_k = p^2/2m = 1.20 \times 10^{-15}\ \text{J}$ .

Key Definitions and Keywords — Momentum and Impulse

Definitions to memorise and the exact keywords mark schemes credit for momentum and impulse answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Linear momentum
Examiner keyword
The product of an object's mass and velocity, $p = mv$ . It is a vector with the unit $\text{kg m s}^{-1}$ and points in the direction of the velocity.
Impulse
Examiner keyword
The product of a force and the time for which it acts, $F\Delta t$ . Impulse equals the change in momentum, $F\Delta t = \Delta p$ . It is a vector with the unit $\text{N s}$ (equivalently $\text{kg m s}^{-1}$ ).
Impulse–momentum relationship
Examiner keyword
The statement that the impulse of the resultant force on an object equals its change in momentum: $F\Delta t = \Delta p = mv - mu$ .
Area under a force–time graph
Examiner keyword
The impulse delivered by a force. For a varying force the impulse must be found as this area (triangle, trapezium or by counting squares), not as a single $F\Delta t$ product.
Newton's second law (momentum form)
Examiner keyword
The resultant force on an object equals the rate of change of its momentum, $F = \Delta p/\Delta t$ . For constant mass this reduces to $\Sigma F = ma$ .
Principle of conservation of linear momentum
Examiner keyword
In a closed system with no external resultant force, the total linear momentum before an interaction equals the total linear momentum after it. In two dimensions each perpendicular component is conserved separately.
Elastic collision
Examiner keyword
A collision in which both total momentum and total kinetic energy are conserved. No kinetic energy is transferred to other forms.
Inelastic collision
Examiner keyword
A collision in which total momentum is conserved but total kinetic energy is not — some kinetic energy is transferred to heat, sound or deformation.
Perfectly inelastic collision
Examiner keyword
An inelastic collision in which the objects stick together and move off with a common velocity, so the maximum possible kinetic energy is lost consistent with momentum conservation.
Kinetic energy
Examiner keyword
The energy an object has because of its motion, $E_k = \tfrac12 mv^2$ . In terms of momentum it can be written $E_k = p^2/2m$ . It is a scalar measured in joules.
Kinetic energy from momentum, $E_k = p^2/2m$
Examiner keyword
The kinetic energy expressed using momentum, obtained by substituting $v = p/m$ into $E_k = \tfrac12 mv^2$ . Widely used in particle physics where momentum is the measured quantity.
Explosion (in mechanics)
An interaction in which a single stationary body separates into parts. The total momentum stays zero, so the fragments' momenta are equal and opposite (and, for two fragments in 1D, the parts recoil in opposite directions).
Recoil
The backward motion of one body when it pushes another forward (e.g. a gun and a bullet). Conservation of momentum makes the two momenta equal and opposite, so the total remains unchanged.
Momentum vector triangle
Examiner keyword
A closed triangle formed by adding momentum vectors tip-to-tail. Used to solve two-dimensional collisions or explosions where momentum before equals momentum after, so the vectors form a closed figure.
Closed (isolated) system
Examiner keyword
A system on which no external resultant force acts, so its total linear momentum is conserved. Only internal forces (which occur in equal-and-opposite pairs) act between the bodies.

Common Mistakes and Misconceptions — Momentum and Impulse

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

✕Ignoring direction (signs) when calculating a change of momentum or applying conservation
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
Momentum is treated as a scalar, so velocities in opposite directions are added with the same sign — especially when a ball rebounds.
How to avoid it
Momentum is a vector. Choose a positive direction and give opposite velocities opposite signs. When an object reverses, $\Delta v = v - (-u)$ , so the speeds add — e.g. $22 - (-9) = 31$ , not $13$ .
✕Assuming kinetic energy is conserved in every collision (or that momentum conservation proves a collision is elastic)
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
Students remember 'momentum is conserved' and wrongly extend it to kinetic energy, or think conserved momentum implies an elastic collision.
How to avoid it
Momentum is conserved in all collisions. Kinetic energy is conserved only in elastic collisions. To classify a collision you MUST compare total KE before and after — a fall in KE means it is inelastic (energy → heat/sound/deformation).
✕Using $F\Delta t$ with a single force value when the force varies (instead of the area under the F–t graph)
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
The formula $F\Delta t$ is memorised without the condition that $F$ must be constant; students read the peak force off the graph and multiply by the total time.
How to avoid it
For a varying force the impulse is the area under the force–time graph. For a triangular graph that area is $\tfrac12 \times \text{base} \times \text{height}$ — half of (peak force × time), not the full rectangle. Count squares for a curved graph.
✕Misusing $E_k = p^2/2m$ — forgetting to square the momentum, or dividing by $m$ instead of $2m$
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
The formula is confused with $p = mv$ , and the factor of 2 and the square are dropped under time pressure.
How to avoid it
Write the equation as $E_k = \dfrac{p^2}{2m}$ and substitute carefully: square $p$ first, then divide by $2m$ . Check units — $\dfrac{(\text{kg m s}^{-1})^2}{\text{kg}} = \text{kg m}^2\text{ s}^{-2} = \text{J}$ .
✕Treating a two-dimensional collision as one-dimensional — adding speeds instead of resolving momentum into components
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
Students apply the familiar 1D 'before = after' equation to angled velocities without resolving, so the perpendicular momentum is ignored.
How to avoid it
In 2D, momentum is conserved separately in two perpendicular directions. Resolve every velocity into components ( $v\cos\theta$ and $v\sin\theta$ ), write one conservation equation per axis, and remember the perpendicular components must sum to the (usually zero) perpendicular momentum before.
✕Giving the wrong unit for impulse or change of momentum, or confusing force with impulse
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
Impulse ( $\text{N s}$ ) and force ( $\text{N}$ ) look similar, and momentum's unit $\text{kg m s}^{-1}$ is unfamiliar.
How to avoid it
Impulse and change of momentum share the unit $\text{N s} = \text{kg m s}^{-1}$ ; a force is in newtons ( $\text{N}$ ) alone. When a question asks for impulse, do not stop at the force — multiply by (or read the area over) the time, and quote $\text{N s}$ .

Momentum and Impulse — frequently asked questions

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

Momentum and Impulse

Short Notes - Momentum and Impulse

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Impulse from force and contact time (2 marks)

2Kinetic energy from momentum, Ek=p2/2mE_k = p^2/2mEk​=p2/2m (2 marks)

3Elastic or inelastic? (3 marks)

4Force from a change of momentum (4 marks)

5Impulse as the area under a force–time graph (4 marks)

6Recoil in an explosion using conservation of momentum (5 marks)

7Two-dimensional collision — resolving momentum (6 marks)

8Momentum vector triangle for an explosion (6 marks)

9Core Practical 9 — force equals rate of change of momentum (6 marks)

Impulse = change in momentum

Impulse as area under a force–time graph

Newton's second law (rate of change of momentum)

Linear momentum

Kinetic energy in terms of momentum

Linear momentum

Impulse

Impulse–momentum relationship

Area under a force–time graph

Newton's second law (momentum form)

Principle of conservation of linear momentum

Elastic collision

Inelastic collision

Perfectly inelastic collision

Kinetic energy

Kinetic energy from momentum, Ek=p2/2mE_k = p^2/2mEk​=p2/2m

Explosion (in mechanics)

Recoil

Momentum vector triangle

Closed (isolated) system

✕Ignoring direction (signs) when calculating a change of momentum or applying conservation

✕Assuming kinetic energy is conserved in every collision (or that momentum conservation proves a collision is elastic)

✕Using FΔtF\Delta tFΔt with a single force value when the force varies (instead of the area under the F–t graph)

✕Misusing Ek=p2/2mE_k = p^2/2mEk​=p2/2m — forgetting to square the momentum, or dividing by mmm instead of 2m2m2m

✕Treating a two-dimensional collision as one-dimensional — adding speeds instead of resolving momentum into components

✕Giving the wrong unit for impulse or change of momentum, or confusing force with impulse

Momentum and Impulse — frequently asked questions

2Kinetic energy from momentum, $E_k = p^2/2m$ (2 marks)

Kinetic energy from momentum, $E_k = p^2/2m$

✕Using $F\Delta t$ with a single force value when the force varies (instead of the area under the F–t graph)

✕Misusing $E_k = p^2/2m$ — forgetting to square the momentum, or dividing by $m$ instead of $2m$