What is the definition of momentum in the context of Edexcel International A Levels Mechanics 1?

In the Edexcel International A Levels Mechanics 1 syllabus, momentum is defined as the product of an object's mass and its velocity. It is a vector quantity, meaning it has both magnitude and direction, and is expressed in units of kg m/s.

How is impulse related to momentum in Mechanics 1?

Impulse is the change in momentum of an object when a force is applied over a period of time. In Mechanics 1, impulse is calculated as the product of the force and the time duration for which it acts. The impulse-momentum theorem states that the impulse on an object is equal to the change in its momentum.

Can you explain the principle of conservation of momentum with an example?

The principle of conservation of momentum states that in a closed system with no external forces, the total momentum before an interaction is equal to the total momentum after the interaction. For example, in a collision between two cars, the sum of their momenta before the collision equals the sum of their momenta after the collision, assuming no external forces like friction are involved.

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

In an elastic collision, both momentum and kinetic energy are conserved. In contrast, in an inelastic collision, momentum is conserved but kinetic energy is not; some of it is transformed into other forms of energy, such as heat or sound. Impulse is involved in both types of collisions as it represents the change in momentum of the objects involved.

How do you calculate the impulse experienced by an object in a given scenario?

To calculate the impulse experienced by an object, you need to know the force applied and the duration of time it acts. Impulse is given by the formula Impulse = Force × Time. Alternatively, if you know the initial and final velocities of the object, impulse can also be calculated as the change in momentum, which is Mass × (Final Velocity - Initial Velocity).

Why is "Adding momenta of particles moving in opposite directions without using signs" a common mistake in Momentum and Impulse?

Why it happens: Treating momentum as a scalar. How to avoid it: Pick a direction as positive. Particles moving the other way have NEGATIVE velocity (and momentum). Write the convention explicitly: 'Take the initial direction of A as positive.' Source: WME01/01 examiner reports — flagged annually

Why is "Computing impulse as $m|v - u|$ when the particle bounces back" a common mistake in Momentum and Impulse?

Why it happens: Forgetting that v should be NEGATIVE after a reversal. How to avoid it: Use signed velocities in I = m(v - u). For a ball that hits a wall at +8 and rebounds at -5: v = -13 (not 3). Then take magnitude at the end.

Why is "Using $m_1 v_1 + m_2 v_2$ on the RHS for a coalescence" a common mistake in Momentum and Impulse?

Why it happens: Reflex to write the standard two-velocity equation. How to avoid it: When particles coalesce, they have a SINGLE velocity. RHS = (m_1 + m_2)v.

Why is "Assuming kinetic energy is conserved in all collisions" a common mistake in Momentum and Impulse?

Why it happens: Generalising from elastic collisions (which are not on the M1 syllabus anyway). How to avoid it: M1 makes NO claim about energy conservation in collisions. Momentum is ALWAYS conserved (assuming no external impulse). KE is conserved only in 'perfectly elastic' collisions — a concept reserved for M2.

Why is "Calculating total momentum after the collision and comparing to a single particle's initial momentum" a common mistake in Momentum and Impulse?

Why it happens: Forgetting to include both particles on both sides. How to avoid it: Conservation is between the TOTAL momentum before and the TOTAL momentum after. Each side includes both particles.

Mechanics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Momentum and Impulse

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Momentum and Impulse Study Notes — Edexcel IAL Mathematics M1 (WME01/01, 2018 spec — 2026 onwards)

Momentum $\mathbf{p} = m\mathbf{v}$ . Impulse $\mathbf{I} = \mathbf{F}t = \Delta\mathbf{p}$ . In a closed system, total momentum is conserved across any collision. Direct (one-dimensional) collisions: handle with signed scalars. M1 does NOT include coefficient of restitution — that's reserved for M2.

At a glance

Momentum: $\mathbf{p} = m\mathbf{v}$ (vector). Units: kg m s $^{-1}$ = N s.
Impulse: $\mathbf{I} = \mathbf{F}t = \Delta\mathbf{p} = m(\mathbf{v} - \mathbf{u})$ (vector).
Conservation: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ for direct collisions.
Coalescence (stick together): $m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$ .
Explosion (from rest): $0 = m_1 v_1 + m_2 v_2$ ; fragments have equal & opposite momenta.
ALWAYS pick a positive direction along the line of motion.
Newton's third law for impulses: $\mathbf{I}_{\text{on A by B}} = -\mathbf{I}_{\text{on B by A}}$ .
M1 does NOT include coefficient of restitution (that's M2).

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

M1 6.1 — Use the definition of momentum and impulse: $\mathbf{p} = m\mathbf{v}$ , $\mathbf{I} = \mathbf{F}t = \Delta\mathbf{p}$ .
M1 6.2 — Apply the principle of conservation of linear momentum to direct collisions.
M1 6.3 — Solve problems involving coalescence (sticking together) and explosions (separation from rest).

Momentum and impulse — definitions

Momentum is mass × velocity. Impulse is force × time = change in momentum.

Momentum of a particle:

$\mathbf{p} = m\mathbf{v}.$

Vector quantity (direction matters).
Units: kg m s $^{-1}$ , equivalent to N s.
A particle at rest has zero momentum.
A heavy slow object can have the same momentum as a light fast one (e.g. $10$ kg at $1$ m/s vs $1$ kg at $10$ m/s — both $10$ N s).

Impulse of a constant force $\mathbf{F}$ acting for time $t$ :

$\mathbf{I} = \mathbf{F}t.$

Vector quantity.
Units: N s.

Impulse-momentum theorem:

$\mathbf{I} = \Delta\mathbf{p} = m(\mathbf{v} - \mathbf{u}).$

Why? From Newton's second law $\mathbf{F} = m\mathbf{a} = m(\mathbf{v} - \mathbf{u})/t$ , so $\mathbf{F}t = m(\mathbf{v} - \mathbf{u})$ .

Practical use: when given a force-time interaction (e.g. ball hitting a wall, golf club striking a ball), use $\mathbf{I} = \Delta\mathbf{p}$ to find the change in velocity or the impulse.

Sign convention: ALWAYS declare a positive direction along the line of motion. A particle that reverses has $\mathbf{v}$ and $\mathbf{u}$ of opposite signs.

Worked example. Ball of mass $0.5$ kg hits wall at $+8$ m/s, rebounds at $-5$ m/s. Impulse on ball:

$I = 0.5(-5 - 8) = -6.5\,\text{N s}.$

Magnitude $6.5$ N s. The negative sign indicates direction (away from the wall).

$\mathbf{p} = m\mathbf{v}$ (vector, kg m s $^{-1}$ ).
$\mathbf{I} = \mathbf{F}t$ (vector, N s).
$\mathbf{I} = m(\mathbf{v} - \mathbf{u}) = \Delta\mathbf{p}$ .
Use signed velocities; magnitude at the end.

Conservation of momentum

Total momentum before = total momentum after, when no external impulse.

Principle. In a closed system (no external impulse during the interaction), the total linear momentum is conserved before and after a collision.

For two particles in a direct (one-dimensional) collision:

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

$u_i$ = velocity of particle $i$ BEFORE the collision (signed).
$v_i$ = velocity of particle $i$ AFTER the collision (signed).
All velocities along the SAME line; sign convention as declared.

Why does it hold? The collision forces are internal to the system. By Newton's third law, the impulse on $A$ from $B$ equals MINUS the impulse on $B$ from $A$ . So $\Delta\mathbf{p}_A = -\Delta\mathbf{p}_B$ , i.e. $\Delta(\mathbf{p}_A + \mathbf{p}_B) = 0$ .

Workflow for a collision problem:

Draw a 'before' and 'after' picture. Mark velocities with arrows.
Declare a positive direction.
List $m_1, u_1, m_2, u_2, v_1, v_2$ with signs.
Apply conservation: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ .
Solve for the unknown.

Sense-check: post-collision speeds should be physical. If particle $A$ caught up with $B$ and they didn't stick, then $v_A \le v_B$ after the collision (otherwise $A$ would penetrate $B$ ).

Worked example. $m_A = 3$ at $5$ m/s; $m_B = 2$ at $-2$ m/s (toward $A$ ). After collision, $v_A = 1$ . Find $v_B$ .

$3 \cdot 5 + 2 \cdot (-2) = 3 \cdot 1 + 2 \cdot v_B$ .
$11 = 3 + 2 v_B \Rightarrow v_B = 4$ m/s (in the original direction of $A$ ).

$B$ has reversed direction — it's now moving in $A$ 's original direction.

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ .
Use signed velocities; declare positive direction.
Holds when no external impulse during the collision.
Sense-check: $v_A \le v_B$ if they don't stick.

Coalescence (stick together)

Combined mass moves at a single velocity. Momentum conserved; KE not.

Setup. Two particles collide and stick together. After the collision, they move as a SINGLE combined mass $m_1 + m_2$ at common velocity $v$ .

Conservation of momentum:

$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \Rightarrow v = \dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2}.$

Worked example. $4$ kg at $6$ m/s hits stationary $2$ kg, stick.

$v = (4 \cdot 6 + 2 \cdot 0)/(4 + 2) = 24/6 = 4\,\text{m s}^{-1}$ .

Kinetic energy? NOT conserved. Before: $\tfrac{1}{2}(4)(6)^{2} = 72$ J. After: $\tfrac{1}{2}(6)(4)^{2} = 48$ J. Lost: $24$ J (converted to heat, sound, deformation).

Why is KE lost? A 'sticking' collision is INELASTIC. There's no spring or elastic property that would store and return the energy. Real-world: car crash where the cars crumple together.

Momentum is ALWAYS conserved (no external impulse), but kinetic energy is conserved ONLY in perfectly elastic collisions — a concept reserved for M2 with the coefficient of restitution.

Coalesce: RHS uses $(m_1 + m_2) v$ .
$v = (m_1 u_1 + m_2 u_2)/(m_1 + m_2)$ .
Momentum conserved; KE not conserved (energy lost to heat/sound).
M1 doesn't ask about KE conservation explicitly.

Explosion (separation from rest)

Initially stationary object splits; fragments have equal & opposite momenta.

Setup. An object of mass $m_1 + m_2$ initially at rest is internally driven apart (explosion, spring release, recoil). After separation, fragment $1$ moves at $v_1$ and fragment $2$ at $v_2$ along the same line.

Conservation of momentum (initial $= 0$ ):

$0 = m_1 v_1 + m_2 v_2 \Rightarrow m_1 v_1 = -m_2 v_2.$

Fragments have equal and opposite momenta. The lighter one moves faster.

Worked example. $5$ kg at rest explodes into $2$ kg at $9$ m/s and $3$ kg at $v_2$ .

$0 = 2 \cdot 9 + 3 \cdot v_2 \Rightarrow v_2 = -6\,\text{m s}^{-1}$ (opposite direction at $6$ m/s).

Kinetic energy? Initial $= 0$ . Final $= \tfrac{1}{2}(2)(9)^{2} + \tfrac{1}{2}(3)(6)^{2} = 81 + 54 = 135$ J. Energy was RELEASED by the explosion (chemical, elastic).

Same form for gun recoil: a bullet fired from a gun at rest has $m_{\text{gun}} v_{\text{gun}} + m_{\text{bullet}} v_{\text{bullet}} = 0$ , so the gun recoils with momentum equal and opposite to the bullet.

Initial momentum = 0 for the stationary system.
$m_1 v_1 + m_2 v_2 = 0$ : equal and opposite momenta.
Lighter fragment is faster.
KE increases (energy released by explosion).

Impulse from a wall / bat / club

Reverse direction $\Rightarrow$ negative $v$ . Subtract carefully.

Setup. A particle moves at $u$ in one direction, strikes a wall (or is hit by a bat), and rebounds at $v$ in the opposite direction.

Impulse on the particle:

$I = m(v - u).$

Crucially: $u$ and $v$ have OPPOSITE SIGNS in your chosen convention.

Worked example. Ball $0.5$ kg hits wall at $+8$ m/s, rebounds at $-5$ m/s.

$I = 0.5 \cdot (-5 - 8) = -6.5$ N s.
Magnitude $6.5$ N s. Direction: opposite to the original motion (i.e. away from the wall).

By Newton's third law, the impulse exerted on the WALL by the ball is $+6.5$ N s (into the wall).

Average force during contact. If the contact lasts time $\Delta t$ :

$F_{\text{avg}} = \dfrac{I}{\Delta t}.$

For a $6.5$ N s impulse over $0.05$ s contact: $F_{\text{avg}} = 130$ N. Larger contact time $\Rightarrow$ smaller force (the basis of crumple zones, airbags, padded gloves).

Reversal: $u$ and $v$ opposite signs.
$I = m(v - u)$ — use signed values; magnitude at end.
$F_{\text{avg}} = I/\Delta t$ when given contact time.
Newton's third: equal & opposite impulse on the other body.

How it’s examined

Momentum and impulse is a regular Q4-Q5 topic on WME01/01, worth $8$ - $10$ marks. Two-particle collision questions are very common, often with a 'find $v$ then find impulse from one particle on the other' two-part structure. Examiners specifically reward (i) declared sign convention, (ii) labelled 'before/after' diagram, (iii) signed substitution into the conservation equation. The Newton's-third-law cross-check ( $I_A = -I_B$ ) is worth a B-mark on multi-step questions. Common error: forgetting that a reversing particle has $v$ of opposite sign — leading to $|v - u| = |v| - |u|$ instead of $|v| + |u|$ in magnitude.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WME01/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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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.

1Impulse changes velocity of a single particle (5 marks, M1)
Core• Adapted from WME01/01 January 2024 Q5• impulse, single particle, change of momentum
Question
A ball of mass $0.5\,\text{kg}$ moving at $8\,\text{m s}^{-1}$ in a straight line strikes a wall and rebounds straight back along the same line at $5\,\text{m s}^{-1}$ . Find the magnitude of the impulse exerted on the ball by the wall.
Step-by-step solution
1. Step 1
  Pick a sign convention. Take the initial direction (toward the wall) as positive. Then $u = +8$ , $v = -5$ .
2. Step 2
  Impulse $\mathbf{I} = \Delta\mathbf{p} = m(v - u)$ .
  $I = 0.5 \cdot (-5 - 8) = 0.5 \cdot (-13) = -6.5\,\text{N s}$
3. Step 3
  Negative sign indicates impulse on the ball points back (away from the wall), as expected — the wall pushed the ball back.
4. Step 4
  Magnitude: $|I| = 6.5\,\text{N s}$ .
Answer
Magnitude of impulse on the ball $= 6.5\,\text{N s}$ .
Examiner tip
B1 for sign convention. M1 for $I = m(v - u)$ . A1 for $-6.5$ . A1 for $6.5$ (magnitude). The most common error is using $|v| - |u| = 5 - 8 = -3$ or $5 + 8 = 13$ without justifying the sign — students must show the direction reversal via signs.
2Direct collision of two particles — both continue forward (8 marks, M1)
Core• Adapted from WME01/01 June 2024 Q4• momentum, conservation, collision
Question
Particle $A$ of mass $3\,\text{kg}$ moves at $4\,\text{m s}^{-1}$ . It collides directly with particle $B$ of mass $2\,\text{kg}$ moving at $1\,\text{m s}^{-1}$ in the same direction. After the collision, $A$ continues in the same direction with speed $2\,\text{m s}^{-1}$ . Find the speed of $B$ after the collision.
Step-by-step solution
1. Step 1
  Take the initial direction of motion as positive. Sketch the situation: $A$ behind, $B$ ahead, both moving in the same direction; $A$ catches up and collides.
2. Step 2
  List initial and final velocities: $u_A = +4$ , $u_B = +1$ , $v_A = +2$ , $v_B = ?$ .
3. Step 3
  Apply conservation of momentum.
  $m_A u_A + m_B u_B = m_A v_A + m_B v_B$
4. Step 4
  Substitute:
  $3 \cdot 4 + 2 \cdot 1 = 3 \cdot 2 + 2 \cdot v_B \Rightarrow 14 = 6 + 2 v_B$
5. Step 5
  Solve for $v_B$ :
  $v_B = 4\,\text{m s}^{-1}$
6. Step 6
  Check: $v_B = 4 > v_A = 2$ , so $B$ is now moving faster than $A$ — physically sensible, no further collision.
Answer
Speed of $B$ after collision $= 4\,\text{m s}^{-1}$ in the same direction.
Examiner tip
B1 for sign convention / diagram. M1 for conservation equation. A1 for the substitution. M1 for solving. A1 for $v_B = 4$ . The check that $v_B > v_A$ post-collision is good practice — examiners sometimes deduct for physically impossible answers (e.g. $B$ moving slower than $A$ ).
3Coalescence — particles stick together (6 marks, M1)
Extended• Adapted from WME01/01 January 2023 Q3• momentum, coalescence, stick together
Question
A particle of mass $4\,\text{kg}$ moves at $6\,\text{m s}^{-1}$ in a straight line. It collides directly with a stationary particle of mass $2\,\text{kg}$ on the same line. After the collision, the two particles coalesce (stick together). Find the common velocity after the collision.
Step-by-step solution
1. Step 1
  Take the initial direction of motion as positive. Initial: $u_A = +6$ , $u_B = 0$ .
2. Step 2
  After coalescing, both move with the same velocity $v$ . Combined mass $= 4 + 2 = 6\,\text{kg}$ .
3. Step 3
  Conservation of momentum:
  $4 \cdot 6 + 2 \cdot 0 = (4 + 2) v$
4. Step 4
  Solve:
  $24 = 6 v \Rightarrow v = 4\,\text{m s}^{-1}$
5. Step 5
  Direction: positive, so $4\,\text{m s}^{-1}$ in the original direction of $A$ .
Answer
Common velocity after coalescing $= 4\,\text{m s}^{-1}$ in the original direction of motion.
Examiner tip
B1 for sign convention. M1 for $(m_1 + m_2)v$ on RHS (coalesce $\Rightarrow$ combined mass). A1 for equation. M1A1 for solving. ECF clean. Note: coalescence is NOT elastic — kinetic energy is lost, but momentum is still conserved.
4Direct collision with one particle reversing direction (10 marks, M1)
Extended• Adapted from WME01/01 June 2023 Q5• momentum, collision, reverse direction
Question
Particles $A$ (mass $3\,\text{kg}$ , moving at $5\,\text{m s}^{-1}$ ) and $B$ (mass $2\,\text{kg}$ , moving at $2\,\text{m s}^{-1}$ toward $A$ ) collide directly. After the collision, $A$ moves at $1\,\text{m s}^{-1}$ in its original direction. (a) Find the velocity of $B$ after the collision. (b) Find the magnitude of the impulse exerted by $A$ on $B$ .
Step-by-step solution
1. Step 1
  Take $A$ 's initial direction as positive. Then $u_A = +5$ , $u_B = -2$ ( $B$ moves toward $A$ , i.e. opposite direction).
2. Step 2
  After: $v_A = +1$ , $v_B = ?$ .
3. Step 3
  Conservation of momentum:
  $3 \cdot 5 + 2 \cdot (-2) = 3 \cdot 1 + 2 \cdot v_B \Rightarrow 15 - 4 = 3 + 2 v_B$
4. Step 4
  Solve:
  $11 = 3 + 2 v_B \Rightarrow v_B = 4\,\text{m s}^{-1}$
5. Step 5
  Positive $v_B$ means $B$ now moves in $A$ 's original direction — $B$ has reversed.
6. Step 6
  (b) Impulse on $B$ = change in $B$ 's momentum.
  $I_B = m_B(v_B - u_B) = 2(4 - (-2)) = 2 \cdot 6 = 12\,\text{N s}$
7. Step 7
  Cross-check by computing impulse on $A$ : $I_A = 3(1 - 5) = -12\,\text{N s}$ . By Newton's third law, $I_A = -I_B$ . Magnitudes equal $12\,\text{N s}$ . ✓
Answer
(a) $v_B = 4\,\text{m s}^{-1}$ in the original direction of $A$ ( $B$ has reversed). (b) Magnitude of impulse $= 12\,\text{N s}$ .
Examiner tip
B1 for sign convention. M1 for conservation. A1 for the equation. M1A1 for $v_B$ . M1 for $I = m\Delta v$ on $B$ . A1 for magnitude. B1 for cross-check using $A$ . The cross-check is highly recommended — Newton's third law for impulses is a common mark-scheme alternative.
5Explosion: stationary object separates into two pieces (8 marks, M1)
Extended• Adapted from WME01/01 January 2024 Q6• momentum, explosion, separation
Question
A stationary object of mass $5\,\text{kg}$ explodes into two fragments. The first fragment has mass $2\,\text{kg}$ and moves at $9\,\text{m s}^{-1}$ in one direction. (a) Find the speed and direction of the second fragment. (b) Comment on whether kinetic energy is conserved.
Step-by-step solution
1. Step 1
  Take the direction of the first fragment's motion as positive. Initial momentum $= 0$ (the object is stationary).
2. Step 2
  Conservation of momentum:
  $0 = 2 \cdot 9 + 3 \cdot v_2$
3. Step 3
  Solve for $v_2$ :
  $v_2 = -\dfrac{18}{3} = -6\,\text{m s}^{-1}$
4. Step 4
  Negative sign means the second fragment moves in the opposite direction at speed $6\,\text{m s}^{-1}$ .
5. Step 5
  (b) Initial KE $= 0$ . Final KE $= \tfrac{1}{2}(2)(9)^{2} + \tfrac{1}{2}(3)(6)^{2} = 81 + 54 = 135\,\text{J} > 0$ . Kinetic energy is NOT conserved — energy was released by the explosion (e.g. from chemical / elastic potential energy).
Answer
(a) Second fragment: $6\,\text{m s}^{-1}$ in the opposite direction. (b) KE is NOT conserved; $135$ J of energy was released by the explosion.
Examiner tip
B1 for sign convention. M1 for conservation with initial momentum $= 0$ . A1 for equation. M1A1 for $v_2$ . B1 for direction. M1A1 for KE comparison. M1 stays consistent — momentum is ALWAYS conserved in collisions and explosions (no external impulse); KE is conserved only in elastic collisions.

Key Formulae — Momentum and Impulse

The formulae you need to memorise for momentum and impulse on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Momentum
$\mathbf{p} = m\mathbf{v}$
$\mathbf{p}$
momentum (kg m s $^{-1}$ or N s)
$m$
mass (kg)
$\mathbf{v}$
velocity (m s $^{-1}$ , vector)
When to use
Defines momentum. Vector: direction matters. Units: kg m s $^{-1}$ are equivalent to N s. NOT in the IAL formula booklet.
Example
Ball of mass $0.5$ kg at $8\,\text{m s}^{-1}$ : $p = 4\,\text{kg m s}^{-1}$ .
Impulse
$\mathbf{I} = \mathbf{F} t = \Delta\mathbf{p} = m(\mathbf{v} - \mathbf{u})$
$\mathbf{I}$
impulse (N s)
$\mathbf{F}$
constant force applied for time $t$
$m, \mathbf{u}, \mathbf{v}$
mass, initial velocity, final velocity
When to use
Connects force-time to change of momentum. Use directly when given a force and duration; use the change-of-momentum form when given initial and final velocities. NOT in the formula booklet.
Example
Ball $0.5$ kg hits wall at $+8$ , rebounds at $-5$ : $I = 0.5(-5 - 8) = -6.5$ N s, magnitude $6.5$ N s.
Conservation of linear momentum
$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
$m_1, m_2$
masses of the two colliding particles
$u_1, u_2$
velocities (with sign) before collision
$v_1, v_2$
velocities (with sign) after collision
When to use
Whenever two particles collide on a smooth surface with no external horizontal impulse. Always state the sign convention along the line of motion. NOT in the formula booklet.
Example
$3$ kg at $5$ vs $2$ kg at $-2$ , after $v_A = 1$ : $15 - 4 = 3 + 2v_B \Rightarrow v_B = 4$ .
Coalescence
$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$
$v$
common final velocity of the combined object
When to use
When particles stick together after collision. The combined mass moves at a single velocity. Momentum conserved; kinetic energy NOT.
Example
$4$ kg at $6$ hits $2$ kg at rest, stick: $24 = 6v \Rightarrow v = 4\,\text{m s}^{-1}$ .
Explosion / separation from rest
$0 = m_1 v_1 + m_2 v_2$
$v_1, v_2$
velocities of the two fragments
When to use
An initially stationary object separates (explodes) into two fragments. Momentum starts at zero, so the two fragments have equal and opposite momenta.
Example
$5$ kg explodes into $2$ kg at $+9$ and $3$ kg at $v_2$ : $0 = 18 + 3v_2 \Rightarrow v_2 = -6$ (opposite direction at $6\,\text{m s}^{-1}$ ).

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 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Momentum
Examiner keyword
The product of mass and velocity: $\mathbf{p} = m\mathbf{v}$ . A vector quantity. SI unit: kg m s $^{-1}$ (equivalent to N s). Conserved in any closed system (no external impulse).
Impulse
Examiner keyword
The product of a force and the time for which it acts: $\mathbf{I} = \mathbf{F}t$ . Equal to the change in momentum it produces: $\mathbf{I} = \Delta\mathbf{p}$ . A vector quantity. SI unit: N s.
Principle of conservation of momentum
Examiner keyword
In a closed system (no external impulse), total linear momentum is constant. For two particles colliding in a straight line: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ .
Direct collision
Examiner keyword
A collision in which both particles move along the same straight line before and after. One-dimensional — can be analysed with signed scalars.
Coalesce
Examiner keyword
Particles stick together after collision and move with a common velocity. Momentum is conserved; kinetic energy is generally NOT conserved (some is converted to heat/sound/deformation).

Common Mistakes and Misconceptions — Momentum and Impulse

The traps other students keep falling into on momentum and impulse questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Adding momenta of particles moving in opposite directions without using signs
WME01/01 examiner reports — flagged annually
Why it happens
Treating momentum as a scalar.
How to avoid it
Pick a direction as positive. Particles moving the other way have NEGATIVE velocity (and momentum). Write the convention explicitly: 'Take the initial direction of $A$ as positive.'
✕Computing impulse as $m|v - u|$ when the particle bounces back
Why it happens
Forgetting that $v$ should be NEGATIVE after a reversal.
How to avoid it
Use signed velocities in $I = m(v - u)$ . For a ball that hits a wall at $+8$ and rebounds at $-5$ : $\Delta v = -13$ (not $3$ ). Then take magnitude at the end.
✕Using $m_1 v_1 + m_2 v_2$ on the RHS for a coalescence
Why it happens
Reflex to write the standard two-velocity equation.
How to avoid it
When particles coalesce, they have a SINGLE velocity. RHS = $(m_1 + m_2)v$ .
✕Assuming kinetic energy is conserved in all collisions
Why it happens
Generalising from elastic collisions (which are not on the M1 syllabus anyway).
How to avoid it
M1 makes NO claim about energy conservation in collisions. Momentum is ALWAYS conserved (assuming no external impulse). KE is conserved only in 'perfectly elastic' collisions — a concept reserved for M2.
✕Calculating total momentum after the collision and comparing to a single particle's initial momentum
Why it happens
Forgetting to include both particles on both sides.
How to avoid it
Conservation is between the TOTAL momentum before and the TOTAL momentum after. Each side includes both particles.

Practice questions

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

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Momentum and Impulse — frequently asked questions

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

Momentum and Impulse

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Impulse changes velocity of a single particle (5 marks, M1)

2Direct collision of two particles — both continue forward (8 marks, M1)

3Coalescence — particles stick together (6 marks, M1)

4Direct collision with one particle reversing direction (10 marks, M1)

5Explosion: stationary object separates into two pieces (8 marks, M1)

Momentum

Impulse

Conservation of linear momentum

Coalescence

Explosion / separation from rest

Momentum

Impulse

Principle of conservation of momentum

Direct collision

Coalesce

✕Adding momenta of particles moving in opposite directions without using signs

✕Computing impulse as m∣v−u∣m|v - u|m∣v−u∣ when the particle bounces back

✕Using m1v1+m2v2m_1 v_1 + m_2 v_2m1​v1​+m2​v2​ on the RHS for a coalescence

✕Assuming kinetic energy is conserved in all collisions

✕Calculating total momentum after the collision and comparing to a single particle's initial momentum

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Momentum and Impulse — frequently asked questions

✕Computing impulse as $m|v - u|$ when the particle bounces back

✕Using $m_1 v_1 + m_2 v_2$ on the RHS for a coalescence