Momentum, Impulse, and the Sign-Convention Trap

~45 min · WME01 · 4.3

WME01 · 4.3 · 45 min

The arithmetic in a momentum question is almost always short — one equation, substituted once. What sinks an otherwise perfect script is a single sign at the very end, and it is not a rare slip: three separate real series (Jan 2020, Oct 2022, Jun 2024) each record the same last mark lost, for the same reason — a candidate reports a velocity's sign where the question asked for a speed, and a speed is never negative.

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

Momentum, impulse, and the one formula spec 4.3 names in words

The momentum of a particle of mass $m$ moving with velocity $v$ is defined as $p = mv$ (spec 4.3). It is a vector: in the one-dimensional problems this unit is confined to — the spec's own guidance for 4.3 states plainly, "Problems will be confined to those of a one-dimensional nature" — that just means momentum carries a sign, the same sign as the velocity producing it, and not merely a size.

The impulse of a force is defined as the change in momentum it produces: $I = mv - mu$ , where $u$ is the velocity before the force acts and $v$ the velocity after. Spec 4.3 names this the impulse-momentum principle, and it is the entire technical content of the topic written as one equation. Everything else in this lesson is either this same equation applied once, to a single particle (an impulse question), or applied twice, once to each of two colliding particles, with the results added together (conservation of momentum).

Neither formula is printed anywhere in the exam room. The M1 section of the real formula booklet — the one Pearson physically issues, Issue 2, January 2021 — reads, in its entirety: "There are no formulae given for M1 in addition to those candidates are expected to know." The spec's own list of what must therefore come from memory names both of these formulae directly: "Momentum = mv", "Impulse = mv − mu". Unlike a Statistics or Further Pure lesson, which can often lean on a reference sheet mid-question, a Mechanics 1 script cannot: the two equations this entire lesson is built from have to already be in the candidate's head.

One explicit scope limit is worth knowing before it costs exam time: spec 4.3's own guidance states, "Knowledge of Newton's law of restitution is not required." Some other Mechanics syllabuses build collision questions around a coefficient of restitution — a number describing how "bouncy" a collision is, linking the speed of separation after to the speed of approach before. WME01 does not examine it. Every collision question on this paper supplies enough information — almost always one particle's velocity after the collision is given directly — for conservation of momentum alone to find everything else that's asked for.

Mechanism

Why momentum is conserved at all — it falls out of Newton's third law, not a separate law

During a direct collision, particle C exerts a force on particle D, and by Newton's third law, D exerts an equal and opposite force back on C, at every instant of contact — the two forces describe the same physical push, seen from its two sides. Because they act for exactly the same duration, the impulses they produce are also equal and opposite: the impulse D receives from C is the negative of the impulse C receives from D. Write that using the impulse-momentum principle applied separately to each particle: impulse on C $= m_C(v_C - u_C)$ , impulse on D $= m_D(v_D - u_D)$ , and "equal and opposite" means $m_C(v_C - u_C) = -m_D(v_D - u_D)$ . Expand both sides and rearrange: $m_C v_C - m_C u_C = -m_D v_D + m_D u_D$ , so $m_C u_C + m_D u_D = m_C v_C + m_D v_D$ — total momentum before the collision equals total momentum after it. Conservation of momentum is not an extra law bolted onto Newton's three; it is exactly what Newton's third law says once it is written in terms of impulse rather than force, applied to two particles instead of one. This derivation also explains, mechanically rather than as a rule to remember, why one shared sign convention has to cover BOTH particles at once: the identity above only holds if $u_C$ , $u_D$ , $v_C$ and $v_D$ are all measured against the same chosen positive direction. Switch which way is "positive" halfway through a working and the two sides of the equation stop describing the same collision.

Diagram — Before and after — a two-particle direct collision, drawn against the sign convention you choose

x-axis: position along the single straight line both particles move on — the positive direction is chosen ONCE, before the equation is written, and every velocity is signed relative to it · y-axis: top = the instant before the collision, bottom = the instant after — not a numerical axis; it exists only to separate the two moments in time on the page

C before the collision: u = 9, moving in the chosen positive direction: Particle C, mass 1 kg, moving at 9 m/s in the direction taken as positive. Drawn as an arrow pointing along the positive direction — the length of the arrow represents the speed, but it is the direction the arrow points, not its length, that the working actually has to keep track of correctly.
D before the collision: u = 0, at rest: Particle D, mass 2 kg, stationary in C's path. A stationary particle contributes mass × 0 = 0 to the total momentum whichever direction is called positive, which is exactly why starting the analysis from D's OWN velocity change, rather than C's, sidesteps a sign question later — see the method comparison further down.
D after the collision: v = +6, same direction as C's original motion: Struck by C, D moves off in the positive direction — consistent with being pushed forward by the impact. This is the value the question gives directly in the worked scenario below.
C after the collision: v = −3, REVERSED direction: The value conservation of momentum actually returns for C's velocity is negative: C has rebounded and is now moving in the direction opposite to the one chosen as positive. This is exactly the number a working must not silently report as a "speed" without first taking its magnitude.

Positive direction, fixed once, before the equation: Here: the direction C is moving in before the collision. Every velocity in the problem — given or found, before or after — is signed relative to this one choice. The choice itself is arbitrary; choosing the opposite direction as positive would flip every sign in the working and produce exactly the same physical answer at the end.
A negative RESULT is not a mistake: Solving the conservation-of-momentum equation for C's velocity correctly returns −3. That number is right, and nothing in the algebra that produced it needs to be redone. What goes wrong, when it does, happens one step later — at the point of reporting it.
Speed = |velocity|, stated with direction in words if needed: "Speed of C = 3 m/s" is the correct final answer to a question that asks for speed. If direction also matters to the answer, it is added as a separate sentence — "3 m/s, in the direction opposite to its original motion" — never folded back into the number as a minus sign.

Common error: Writing the final answer to "find the speed of C after the collision" as −3 m/s, copied straight from the conservation-of-momentum working.

Correct: Take the magnitude of the signed velocity the equation returned: speed = |−3| = 3 m/s. The equation was correct to produce a negative number; a speed is simply never allowed to be reported as one.

examiner-report · Jan 2020 · Q1

Worked, in full

The velocity — and then the speed — of C after a direct collision with a stationary particle

01
Draw the situation and fix a positive direction before writing anything down. Two particles, C (mass 1 kg) and D (mass 2 kg), move in the same straight line on a smooth horizontal surface. C is moving at 9 m/s; D is at rest. Take the direction C is moving in as positive — an arbitrary choice, but a required one, made once, before the collision itself is even discussed.
Earns: Nothing gradeable yet, but this is precisely what a real 'M' mark for Mechanics rewards a moment later: the general marking guidance for M marks states they are usually awarded 'for the application of some mechanical principle to produce an equation... applying the conservation of momentum principle etc.' — and that principle can only be applied correctly once every velocity feeding into it has a sign fixed against a stated convention.
02
Write down every velocity, signed, relative to that one convention: $u_C = 9$ , $u_D = 0$ . After the collision, D moves off with velocity $v_D = 6$ (given, in the same positive direction — D is pushed forward by the impact). C's velocity after the collision, $v_C$ , is the unknown.
Earns: M1 — sets up the problem with correctly signed values for the three velocities already known.
03
Apply conservation of momentum for the two particles: total momentum before the collision equals total momentum after it. $m_C u_C + m_D u_D = m_C v_C + m_D v_D$ , so $1(9) + 2(0) = 1(v_C) + 2(6)$ .
Earns: M1 — applies the conservation-of-momentum principle itself, $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ : the named spec-4.3 content, not a bespoke equation invented for this one scenario.
04
Solve for the unknown. $9 = v_C + 12$ , so $v_C = 9 - 12 = -3$ .
Earns: A1 — correct signed velocity. The value genuinely is negative; there is no step above where the sign 'should' instead have come out positive.
05
Answer the question actually asked. If the question wants the velocity of C, $-3$ m/s is a complete answer on its own. If, as the real question this scenario is modelled on does, it instead asks for the SPEED of C after the collision, the answer is $|-3| = 3$ m/s — positive, because a speed is a magnitude, and a magnitude cannot be negative whatever sign the underlying velocity carries.
Earns: A1 — final answer positive, 3 m/s: exactly the mark real candidates are recorded losing when they instead copied −3 m/s straight down against a 'find the speed' instruction.

Source — Examiner report, Jan 2020

"the most common error was in not taking account of the direction of the velocity and so not producing a positive answer as required for the 'speed' of P"

In your own words

In one sentence: why can a velocity inside a conservation-of-momentum equation come out negative, while an answer the question calls a "speed" never can?

Same question, every valid method

Continuing the collision above: C (mass 1 kg, initial velocity 9 m/s) collides directly with the stationary D (mass 2 kg). After the collision D moves off at 6 m/s in the direction C was originally travelling, and (from the worked chain above) C moves off at 3 m/s in the reverse direction. Find the magnitude of the impulse exerted on D by C in the collision. (VERIDIAN-original question, continuing the scenario above — not a reproduction of any real WME01 question. The research bank documents examiner commentary on real Jan 2020/Oct 2022/Jun 2024 momentum questions, not their numeric stems, so no real numbers exist here to reproduce.)

2 valid methods · every one reaches 12 N s (equivalently, 12 kg m/s), in the direction C was originally travelling · 2 marks available

01
$I_{\text{on }D} = m_D(v_D - u_D) = 2(6 - 0) = 12$
Method mark for applying the impulse-momentum principle, I = mv − mu, to a single particle. D's own initial velocity is already zero, so the working needs only D's own two numbers — nothing about C is used at all.
M1
02
Magnitude of the impulse exerted on D: 12 N s.
Accuracy mark, reached in a single substitution with no further reasoning required.
A1

Always the faster route when one particle starts at rest, or simply has the plainer known velocities: the impulse ON that particle is one substitution into I = mv − mu, using only that particle's own mass and its own two velocities. Nothing about the other particle, and no invocation of Newton's third law, is needed anywhere in the working.

Named traps

velocity-sign-carried-into-speed-answer: The single strongest, most independently confirmed trap anywhere in the WME01 examiner-report record — verified across three separate series, not one. Jan 2020: "the most common error was in not taking account of the direction of the velocity and so not producing a positive answer as required for the 'speed' of P." Oct 2022: "candidates need to be reminded that the final answer needed to be positive as speed was required." Jun 2024: "the final mark was often lost because it had been left as negative." All three describe the same failure: the working is correct up to and including a genuinely negative velocity, and the final mark is lost only by copying that sign into an answer the question specifically called a speed. The fix costs one line: take the magnitude, and state the direction in words if the question wants it too.
harder-particle-chosen-for-impulse: Confirmed on a real WME01 impulse question: "the majority of responses used particle A to find the impulse, rather than using particle B which was much easier since it started at rest" (Jun 2024, Q1). This isn't a correctness error — both routes reach the same right answer — but it is a genuine, examiner-documented efficiency cost: computing the impulse via the harder particle first, then having to invoke Newton's third law to flip the result onto the particle actually asked about, is an extra step with its own extra chance to drop a sign, for no extra credit. Before substituting anything, check which of the two particles in the question has the simpler known velocities (often, but not only, the one that started at rest) and start there.

Retrieval — with feedback on every choice

Question 1

3 marks

Particles E (mass 4 kg, speed 3 m/s) and F (mass 6 kg, speed 4 m/s) move towards each other along the same straight line on a smooth horizontal surface and collide directly, coalescing into a single particle. Taking the direction E was moving as positive, what is the velocity of the combined particle after the collision?

Question 2

2 marks

A question asks for the magnitude of the impulse exerted on particle H, which was at rest before colliding directly with G (already moving). Which single method reaches that answer in one substitution into I = mv − mu, with no further reasoning step needed?

Question 3

2 marks

Which of the following, unlike on some other exam boards' Mechanics syllabuses, is explicitly NOT required knowledge for a WME01 momentum/impulse question (spec 4.3)?

Question 4

1 mark

Particles J (mass 5 kg) and K (mass 3 kg) collide directly. After the collision, J's velocity is $-1$ m/s (taking J's original direction of motion as positive). A question part asks: "Find the speed of J after the collision." What is the correct final answer?

Reference — not a study method, a lookup

Momentum p = mv. Impulse I = mv − mu (the impulse-momentum principle). Neither is in the M1 formula booklet — memorise both.
Conservation, two particles colliding directly: m1u1 + m2u2 = m1v1 + m2v2. Every velocity is SIGNED, not a bare speed.
Fix a positive direction ONCE, before writing the equation. Every velocity — given or found — is signed relative to it.
A negative result is not an error. Reporting it as a 'speed' is: speed = |velocity|, never negative.
For an impulse question, start with whichever particle has the simpler known velocities (often the one at rest) — one substitution, no Newton's-third-law flip needed.
Restitution is NOT required at M1 (spec 4.3). Collision problems are confined to one dimension.

Not affiliated with or endorsed by Pearson Edexcel. Every examiner-report quotation in this lesson — the three-series sign-convention trap (Jan 2020 Q1, Oct 2022 Q1, Jun 2024 Q1) and the particle-choice efficiency note (Jun 2024 Q1) — is verified word-for-word against WME01-verified-facts.md, itself checked against the actual downloaded PDF text of each report rather than a secondary summary. The formulae used (momentum = mv, impulse = mv − mu, conservation of momentum for two particles colliding directly) are spec 4.3's own named M1 content, confirmed absent from the real exam formula booklet and therefore taught here as content to memorise rather than as a citation needing its own separate source. The numeric collision scenarios running through the worked chain, method comparison, prequestions and MCQs (particles C and D; E and F; J and K) are all VERIDIAN-original: the research bank documents what three real series' examiner reports SAID about the errors candidates made on their own Q1 momentum questions, not the numeric stems of those real questions, so there is nothing genuine to reproduce. Every scenario was checked for physical consistency before being written in — total momentum balances exactly in each one, and the main C/D scenario happens to conserve kinetic energy exactly (40.5 J before and after), confirming the 'rebound' it describes is a genuinely possible collision rather than an equation that merely balances on paper.

Question 13 marks

$-1.2$ m/s
Correct. F moves towards E, so F's velocity is $-4$ relative to the chosen convention: $4(3) + 6(-4) = 12 - 24 = -12$ . The combined mass is $10$ kg, so $v = -12 \div 10 = -1.2$ m/s — the combined particle moves off in the direction F was originally travelling, since F's momentum outweighed E's.
B $3.6$ m/s
This treats both speeds as positive: $4(3) + 6(4) = 36$ , giving $3.6$ . That's only valid if E and F move the SAME way — but the question states they move towards each other, which is exactly the information a signed velocity exists to carry.
C $1.2$ m/s
This assigns the negative sign to E instead of F: $4(-3) + 6(4) = -12 + 24 = 12$ , divided by the combined mass gives $1.2$ . The question fixes E's OWN direction as positive, so E's velocity is $+3$ ; it is F, moving towards E, that takes the negative sign — not the other way round.
D $0.2$ m/s
This pairs each particle's mass with the OTHER particle's velocity: $6(3) + 4(-4) = 18 - 16 = 2$ , divided by the combined mass $10$ . Each particle's own mass has to be multiplied by its own velocity — swapping them silently changes which particle's momentum is being computed.

Traps tested: Opposing direction not signed negative · Sign assigned to wrong particle · Masses and velocities mismatched

Question 22 marks

Apply I = mv − mu to H itself, using H's own velocity before (zero) and after the collision.
Correct. The impulse exerted ON H is, by definition, H's own change in momentum — H's own mass and its own two velocities are all the formula needs. H starting at rest also removes one term from the arithmetic entirely.
BApply I = mv − mu to G, using G's velocities before and after, and take that number as the final answer directly.
This computes the impulse exerted ON G, not on H — Newton's third law says the two are equal in magnitude but OPPOSITE in sign, so taking G's result unchanged gives an answer with the wrong sign for the quantity actually asked about.
CApply I = mv − mu to G, then flip the sign using Newton's third law to obtain the impulse on H.
This is a genuinely valid route to the correct answer — but it takes two steps, not one: a substitution, and then a separate, correctly-signed application of Newton's third law. It answers a different question from the one asked (which method needs no extra step).
DIt cannot be done in a single substitution for either particle, since both G and H are affected by the same collision.
Being part of the same collision doesn't stop H's own momentum change from being computable directly from H's own two velocities — that's exactly what option A does, in one line.

Traps tested: Impulse on other particle used without sign flip · Extra step not recognised as extra · Overclaims symmetry between particles

Question 32 marks

Which of the following, unlike on some other exam boards' Mechanics syllabuses, is explicitly NOT required knowledge for a WME01 momentum/impulse question (spec 4.3)?

Newton's law of restitution / a coefficient of restitution
Correct. Spec 4.3's own guidance states directly, "Knowledge of Newton's law of restitution is not required." Every collision question on this paper supplies enough information — almost always one particle's velocity after the collision — for conservation of momentum alone to solve it.
BThe impulse-momentum principle, I = mv − mu
This IS required — it is the formula spec 4.3 names in words, and the entire technical content of this lesson is built from it.
CConservation of momentum for two particles colliding directly
This IS required — it is named explicitly in spec 4.3's own wording ("the principle of conservation of momentum applied to two particles colliding directly") and is the second half of this lesson's content.
DCorrectly assigning the sign of a velocity relative to a chosen positive direction
This IS required, and it's the single most costly skill in the whole topic to get wrong — it's the specific discipline behind the strongest verified trap in this lesson (losing the final mark for a negative 'speed').

Traps tested: Required content mistaken for excluded

Question 41 mark

$1$ m/s
Correct. Speed = |velocity| = |−1| = 1. This value doesn't need conservation of momentum or K's mass at all — the velocity of J is already given; the only step is taking its magnitude, because the question asked for a speed, not a velocity.
B $-1$ m/s
The same trap as the worked example above, in its simplest possible form: the velocity was GIVEN as negative, and that sign is copied straight into an answer the question called a speed.
C $1$ m/s, but only if J's direction has genuinely reversed — otherwise $-1$ m/s
Speed is |velocity| unconditionally; it doesn't have a "reversed" case and a "not reversed" case with different rules. A negative velocity always means the same thing for the corresponding speed: take the magnitude.
DCannot be found without also knowing K's velocity after the collision
J's speed depends only on J's own velocity, which is already given. K's velocity would be needed to check conservation of momentum or to answer a question ABOUT K — it adds nothing to a question asking only for J's speed.

Traps tested: Velocity sign carried into speed answer · Speed treated as conditionally signed · Unnecessary dependency assumed

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jan 2020 · Q1 — cited directly in this lesson

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Select International Advanced Level → Mathematics → any series, then look for WME01.

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Up next

Friction — One Unified Model for Equilibrium and Motion on a Rough Plane

The WME01 spec states friction as two separate guidance lines under two separate sections — "F = μR when a particle is moving" (4.4) and "F ≤ µR in a situation of equilibrium" (5.3) — and it is entirely possible to revise them as two unrelated facts. Pearson's own question-writing doesn't: a particle resting on a rough plane and a particle accelerating down the same plane are the same scenario, split only by whether the resultant force is zero or not, and the single most heavily reported error on this whole topic — getting the friction force's *direction* wrong — costs marks in both versions equally, because neither equation says anything at all about which way F points.

50 min

Momentum, Impulse, and the Sign-Convention Trap

Momentum, impulse, and the one formula spec 4.3 names in words

Why momentum is conserved at all — it falls out of Newton's third law, not a separate law

The velocity — and then the speed — of C after a direct collision with a stationary particle

Deep dive

Practice this for real