What is the definition of impulse in the context of Mechanics 2?

In Mechanics 2, impulse is defined as the change in momentum of an object when a force is applied over a short time interval. It is a vector quantity and is calculated as the product of the force and the time duration over which the force acts. Impulse is measured in Newton-seconds (Ns) and is crucial in analyzing collisions.

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 solving collision problems in Mechanics 2, as it allows students to calculate unknown velocities or masses of colliding objects.

What is the difference between elastic and inelastic collisions?

In Mechanics 2, an elastic collision is one where both momentum and kinetic energy are conserved. In contrast, an inelastic collision conserves momentum but not kinetic energy, as some energy is transformed into other forms, such as heat or sound. Understanding these differences is vital for solving problems related to impulses and collisions in Edexcel International A Levels.

How do you calculate the impulse experienced by an object during a collision?

To calculate the impulse experienced by an object during a collision, you need to determine the change in momentum of the object. This is done by subtracting the initial momentum (mass times initial velocity) from the final momentum (mass times final velocity). The formula is: Impulse = m(v_f - v_i), where m is the mass, v_f is the final velocity, and v_i is the initial velocity.

Why is understanding impulses and collisions important in the Edexcel International A Levels curriculum?

Understanding impulses and collisions is crucial in the Edexcel International A Levels curriculum because it provides students with a deeper insight into how forces affect motion. This knowledge is not only foundational for advanced studies in physics and engineering but also enhances problem-solving skills applicable in real-world scenarios, such as vehicle safety design and sports dynamics.

Why is "Sign confusion in momentum and restitution equations" a common mistake in Impulses and collisions?

Why it happens: Students forget to assign signs consistently. How to avoid it: Declare positive direction at the start (typically the direction of one particle's initial motion). Apply signs consistently to ALL velocities (negative for opposite direction). After the algebra, check: positive result = same direction; negative = reversed. Source: WME02/01 examiner reports — annual

Why is "Writing the restitution equation as $v_A - v_B = e(u_A - u_B)$" a common mistake in Impulses and collisions?

Why it happens: Memorisation slip — students miss that separation is v_B - v_A, not v_A - v_B, when A was behind B before. How to avoid it: Restitution is ALWAYS separationapproach = e with both quantities positive. Set it up using the physical interpretation: 'after the collision, the gap between them opens at speed v_B - v_A'.

Why is "Using 'conservation of kinetic energy' instead of conservation of momentum" a common mistake in Impulses and collisions?

Why it happens: Energy conservation is the dominant theme in work-energy questions. How to avoid it: KE is conserved ONLY for e = 1 (perfectly elastic). For any e < 1, KE is lost. Momentum is ALWAYS conserved in a collision (provided no external impulses).

Why is "Computing wall impulse as $m v$ or $m u$ instead of $m(v - u)$" a common mistake in Impulses and collisions?

Why it happens: Treating impulse as 'final momentum' rather than 'change of momentum'. How to avoid it: Impulse is ALWAYS p = m(v - u). For wall collisions, v and u have opposite signs, so |I| = m(|u| + |v|) = m u (1 + e) for a perpendicular hit. Source: WME02/01 examiner reports — frequent

Why is "Not checking for further collisions in multi-particle problems" a common mistake in Impulses and collisions?

Why it happens: Once the algebra gives velocities, students move on. How to avoid it: After each collision, check (i) whether the rear particle is now moving faster than the front (would catch up — another collision), (ii) whether any particle has rebounded backward toward an earlier particle (could collide again). Make the check explicit.

Why is "Omitting units on impulse (or quoting kg m/s instead of N s)" a common mistake in Impulses and collisions?

Why it happens: Both are correct SI but the convention is N s. How to avoid it: Quote N s on impulse problems. They're dimensionally equivalent (1 N s = 1 kg m s^-1), but N s is the conventional unit at IAL.

Mechanics 2Edexcel International A Levels Mathematics (XMA01-YMA01)

Impulses and collisions

Q: Why is "Computing wall impulse as $m v$ or $m u$ instead of $m(v - u)$" a common mistake in Impulses and collisions?

Why it happens: Treating impulse as 'final momentum' rather than 'change of momentum'. How to avoid it: Impulse is ALWAYS p = m(v - u). For wall collisions, v and u have opposite signs, so |I| = m(|u| + |v|) = m u (1 + e) for a perpendicular hit. Source: WME02/01 examiner reports — frequent

Q: Why is "Not checking for further collisions in multi-particle problems" a common mistake in Impulses and collisions?

Why it happens: Once the algebra gives velocities, students move on. How to avoid it: After each collision, check (i) whether the rear particle is now moving faster than the front (would catch up — another collision), (ii) whether any particle has rebounded backward toward an earlier particle (could collide again). Make the check explicit.

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Impulses and Collisions Study Notes — Edexcel IAL Mathematics M2 (WME02/01, 2018 spec — 2026 onwards)

Impulse $=$ change of momentum. Collisions: conservation of momentum + Newton's law of restitution ( $e =$ separation/approach). Collisions with walls, successive collisions, kinetic energy loss in inelastic collisions.

What you’ll learn

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

M2 5.1 — Calculate the impulse acting on a particle and link it to change of momentum.
M2 5.2 — Apply conservation of linear momentum to direct collisions between two particles.
M2 5.3 — Use Newton's experimental law (coefficient of restitution) for direct impacts.
M2 5.4 — Solve problems involving successive collisions and loss of kinetic energy.

Impulse and momentum

Impulse is change of momentum. Units: N s.

Momentum. A particle of mass $m$ at velocity $\mathbf{v}$ has momentum $\mathbf{p} = m\mathbf{v}$ . It's a vector — direction matters.

Impulse. When a force $\mathbf{F}$ acts on a particle for time $t$ , it changes the momentum:

$\mathbf{I} = \Delta\mathbf{p} = m\mathbf{v} - m\mathbf{u} = \int_{0}^{t} \mathbf{F}\,dt'.$

For a constant force, $\mathbf{I} = \mathbf{F} t$ .

Units. SI units of impulse: N s = kg m s $^{-1}$ . The two are dimensionally identical; IAL convention is N s.

Worked example. Particle, $m = 0.5$ kg, $\mathbf{u} = 3\mathbf{i} - 2\mathbf{j}$ m/s, $\mathbf{v} = \mathbf{i} + 4\mathbf{j}$ m/s.

$\mathbf{I} = m(\mathbf{v} - \mathbf{u}) = 0.5(\mathbf{i} + 4\mathbf{j} - 3\mathbf{i} + 2\mathbf{j}) = 0.5(-2\mathbf{i} + 6\mathbf{j}) = -\mathbf{i} + 3\mathbf{j}$ N s.
$|\mathbf{I}| = \sqrt{1 + 9} = \sqrt{10}$ N s.

1D shortcut. With signs: impulse = $m(v - u)$ , with $u$ and $v$ signed quantities.

Why useful. Impulse short-circuits the need to know the force or the time individually. For collisions (very large force, very short time), we just compute $\Delta\mathbf{p}$ and call it the impulse delivered.

$\mathbf{p} = m\mathbf{v}$ — momentum, vector.
$\mathbf{I} = \Delta\mathbf{p}$ — impulse equals change of momentum.
$\mathbf{I} = \mathbf{F} t$ for constant force.
Units: N s.

Conservation of momentum in collisions

No external impulse on the system ⇒ total momentum is conserved.

Statement. For two particles colliding on a smooth line (no external impulses during the collision):

$m_A u_A + m_B u_B = m_A v_A + m_B v_B.$

Why. During the collision, the only forces are the equal-and-opposite contact forces (Newton's third law). They produce equal-and-opposite impulses on the two particles, so the total impulse on the pair is zero — hence total momentum is unchanged.

Sign convention. Pick a positive direction (typically the direction of one particle's initial motion). Apply signs consistently. If a particle moves in the negative direction, write a negative number.

Worked example. $A$ (2 kg) at $+6$ m/s; $B$ (3 kg) at $-4$ m/s. After: $v_A$ , $v_B$ .

$2(6) + 3(-4) = 12 - 12 = 0$ — total momentum before is zero.
So $2 v_A + 3 v_B = 0$ ⇒ relation between final velocities.

This is one equation in two unknowns. We need a second equation — restitution.

Smooth surface assumption. 'Smooth' = no friction acting horizontally during the collision. If the surface is rough, the friction force is finite (not impulsive), so over the collision time $t \to 0$ , the friction impulse $F_{\text{fric}} t \to 0$ . Momentum is still conserved across the collision instant. Friction matters BEFORE and AFTER the collision but not DURING.

$\sum m \mathbf{u} = \sum m \mathbf{v}$ .
Vector equation — apply component-by-component in 2D.
Sign matters in 1D.
Friction does not impede conservation during the collision.

Newton's law of restitution

$e = \dfrac{\text{speed of separation}}{\text{speed of approach}}$ , between $0$ and $1$ .

Statement. For a direct collision between two particles along a line, the speed of separation equals $e$ times the speed of approach:

$v_B - v_A = e(u_A - u_B),$

assuming $A$ was behind $B$ before the collision (so $u_A > u_B$ for the particles to be approaching each other). The signed difference $v_B - v_A$ is the separation speed after.

Range of $e$ .

$e = 0$ : perfectly inelastic. Particles coalesce — same final velocity.
$0 < e < 1$ : partially elastic. KE is lost.
$e = 1$ : perfectly elastic. KE is conserved (idealisation).

The two equations. Conservation of momentum and the restitution equation give two linear equations in $v_A, v_B$ . Solve simultaneously.

Worked example. $A$ (2 kg) at $+6$ , $B$ (3 kg) at $-4$ , $e = 0.5$ .

Momentum: $2 v_A + 3 v_B = 0$ (from above).
Restitution: $v_B - v_A = 0.5 \cdot (6 - (-4)) = 5$ .
From restitution: $v_B = v_A + 5$ . Substitute: $2 v_A + 3(v_A + 5) = 0$ , so $5 v_A = -15$ , $v_A = -3$ , $v_B = 2$ m/s.

After the collision, $A$ has reversed direction (now moving left at 3 m/s); $B$ has reversed and moves right at 2 m/s.

Coalescence shortcut. If $e = 0$ , particles share a common final velocity $v$ given by momentum alone: $v = (m_A u_A + m_B u_B)/(m_A + m_B)$ .

Restitution: $v_B - v_A = e(u_A - u_B)$ .
Pair with momentum conservation — two equations, two unknowns.
$e = 0$ : coalesce. $e = 1$ : elastic. In between: partially elastic.
Coalescence: common final velocity $=$ weighted-average initial velocity.

Collisions with a fixed wall

Perpendicular: $v = -e u$ . Oblique: parallel component unchanged.

Perpendicular impact. A particle of mass $m$ strikes a smooth fixed wall with velocity $u$ perpendicular to the wall. The wall has infinite mass (immovable), so:

Momentum: not useful (the wall's momentum change is unmeasurable; treat the particle alone).
Restitution: speed of separation $= e \cdot$ speed of approach, with the wall stationary. So $|v| = e|u|$ , with $v$ in the opposite direction.

In signed form (with the direction toward the wall positive):

$v = -e u.$

Impulse from the wall. $|I| = |m(v - u)| = m|u|(1 + e)$ .

Worked example. $m = 0.2$ kg, $u = 15$ m/s, $e = 0.7$ .

After: $|v| = 0.7 \cdot 15 = 10.5$ m/s, opposite direction.
$|I| = 0.2(15 + 10.5) = 0.2 \cdot 25.5 = 5.1$ N s.

Oblique impact (rare in M2 but worth knowing). Velocity has a component perpendicular to the wall ( $v_{\perp}$ ) and a component parallel ( $v_{\parallel}$ ). Only $v_{\perp}$ is affected by the collision: $v_{\perp, \text{after}} = -e v_{\perp, \text{before}}$ . The parallel component is unchanged (smooth wall = no friction).

Speed after oblique impact. $|v_{\text{after}}| = \sqrt{(e v_{\perp})^{2} + v_{\parallel}^{2}}$ .

Perpendicular wall hit: $v = -eu$ .
Impulse from wall: $|I| = mu(1 + e)$ .
Oblique: only perpendicular component affected.
Smooth wall: parallel component unchanged.

Successive collisions

Multiple particles in a row. Apply momentum + restitution at each collision in turn.

Setup. Three particles $A$ , $B$ , $C$ on a smooth line in that order. $A$ projected toward $B$ ; $B$ , $C$ initially at rest. Coefficients of restitution $e_1$ (between $A$ and $B$ ) and $e_2$ (between $B$ and $C$ ).

Procedure.

Solve collision $A$ - $B$ using momentum + restitution. Get $v_A$ , $v_B$ post-collision-1.
If $v_B > 0$ (i.e. $B$ now moves toward $C$ ), solve collision $B$ - $C$ similarly.
Check for further collisions. Does $A$ catch up to $B$ again? Is the new $v_B$ (after the second collision) less than $v_C$ ? If so, no more $B$ - $C$ collision. If new $v_B$ is negative and exceeds $v_A$ in magnitude, $A$ and $B$ collide again.

Worked example. $A$ (1 kg, $u_A = 10$ ), $B$ (2 kg, rest), $C$ (3 kg, rest); $e_1 = 0.5$ , $e_2 = 0.6$ .

Collision 1: $v_A + 2v_B = 10$ , $v_B - v_A = 5$ . Solve: $v_A = 0$ , $v_B = 5$ .
Collision 2: $2(5) = 2 v_B' + 3 v_C$ , $v_C - v_B' = 0.6 \cdot 5 = 3$ . Solve: $v_B' = 0.2$ , $v_C = 3.2$ .
Check: $v_B' = 0.2 < v_C = 3.2$ , so no further $B$ - $C$ collision. And $v_B' = 0.2 > v_A = 0$ , so $B$ moves away from $A$ — no further $A$ - $B$ collision.

Final velocities: $v_A = 0$ , $v_B = 0.2$ , $v_C = 3.2$ m/s.

Common variant: catch-up. After collision 1, $A$ is left at rest and $B$ moves at some speed. After collision 2, $B$ may bounce back at a speed exceeding $A$ 's. Then a third collision occurs between $A$ and $B$ . The procedure repeats.

Solve each collision in time-order.
Use momentum + restitution at each.
Always check whether the next collision happens.
Make the check explicit in the answer.

Kinetic energy lost in a collision

Compute KE before and KE after; subtract. Always $\geq 0$ .

Computation.

$\Delta\text{KE} = \text{KE}_{\text{before}} - \text{KE}_{\text{after}} = \sum \tfrac{1}{2}m_i u_i^{2} - \sum \tfrac{1}{2}m_i v_i^{2}.$

Worked example. Collision 1 from earlier: 2 kg at $6$ , 3 kg at $-4$ , becoming $-3$ and $2$ .

KE before: $\tfrac{1}{2}(2)(36) + \tfrac{1}{2}(3)(16) = 36 + 24 = 60$ J.
KE after: $\tfrac{1}{2}(2)(9) + \tfrac{1}{2}(3)(4) = 9 + 6 = 15$ J.
KE lost: $60 - 15 = 45$ J.

Elastic check. For $e = 1$ , $\Delta\text{KE}$ should be zero. For $e = 0$ (coalescence), $\Delta\text{KE}$ should be maximum.

Worked example — coalescence. 4 kg at $5$ m/s hits 6 kg at rest, $e = 0$ .

Final $v = 20/10 = 2$ m/s (common).
KE before: $\tfrac{1}{2}(4)(25) = 50$ J. KE after: $\tfrac{1}{2}(10)(4) = 20$ J.
KE lost: $30$ J.

Why the loss. Energy is converted to heat, sound, deformation. M2 doesn't ask where the energy goes — just to compute the loss.

$\Delta\text{KE} =$ KE before $-$ KE after.
$\Delta\text{KE} \geq 0$ always; $= 0$ only for $e = 1$ .
Maximum loss at $e = 0$ (coalescence).
Compute KE term-by-term; sum signed velocities squared.

How it’s examined

Impulses and collisions appears on every WME02/01 paper as a $10$ - $14$ -mark question, often Q6 or Q7. Typical structure: (a) momentum + restitution for a direct collision, (b) impulse on one of the particles, (c) KE lost. The harder variant is the three-particle successive collisions (June 2023, Q8). Examiners aggressively check sign convention and the 'no further collision' verification step. Wall problems and impulse-vector problems also recur. A* candidates should target full marks; the methodology is mechanical once memorised, with the bookkeeping (signs, second collision check) being the discriminator.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WME02/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Impulses and collisions

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

1Direct collision: find velocities and KE lost (8 marks, M2)
Core• Adapted from WME02/01 January 2024 Q6• collision, momentum, restitution
Question
Two particles $A$ (mass $2$ kg) and $B$ (mass $3$ kg) move toward each other on a smooth horizontal line. Before collision, $A$ moves at $6$ m s $^{-1}$ to the right and $B$ at $4$ m s $^{-1}$ to the left. The coefficient of restitution between them is $e = 0.5$ . Find (a) the velocities after the collision, (b) the kinetic energy lost.
Step-by-step solution
1. Step 1
  Take right as positive. Before: $u_A = +6$ , $u_B = -4$ . After: $v_A$ , $v_B$ (unknowns, both positive-direction).
2. Step 2
  Conservation of momentum: $m_A u_A + m_B u_B = m_A v_A + m_B v_B$ .
  $2(6) + 3(-4) = 2 v_A + 3 v_B \Rightarrow 12 - 12 = 2 v_A + 3 v_B \Rightarrow 2 v_A + 3 v_B = 0 \;\;(i)$
3. Step 3
  Newton's law of restitution: $v_B - v_A = e(u_A - u_B)$ .
  $v_B - v_A = 0.5(6 - (-4)) = 0.5 \cdot 10 = 5 \;\;(ii)$
4. Step 4
  From (ii): $v_B = v_A + 5$ . Substitute into (i): $2 v_A + 3(v_A + 5) = 0$ , so $5 v_A = -15$ , $v_A = -3$ m/s. Then $v_B = -3 + 5 = 2$ m/s.
5. Step 5
  Interpretation: $A$ now moves at $3$ m/s to the LEFT (rebounded). $B$ moves at $2$ m/s to the RIGHT.
6. Step 6
  (b) KE before: $\tfrac{1}{2}(2)(6)^{2} + \tfrac{1}{2}(3)(4)^{2} = 36 + 24 = 60$ J. KE after: $\tfrac{1}{2}(2)(3)^{2} + \tfrac{1}{2}(3)(2)^{2} = 9 + 6 = 15$ J.
7. Step 7
  KE lost = $60 - 15 = 45$ J.
Answer
(a) $v_A = -3$ m/s (i.e. $3$ m/s to the left), $v_B = 2$ m/s (to the right). (b) $45$ J of KE lost.
Examiner tip
(a) M1 for momentum equation. A1 for (i). M1 for restitution equation with correct $\sigma$ convention. A1 for (ii). M1 for solving simultaneously. A2 for both velocities. (b) M1 for KE before AND after. A1 for $45$ J. Sign convention is the marker discriminator: students who choose 'left positive for $B$ ' must keep it consistent throughout.
2Impulse on a particle in 2D (5 marks, M2)
Core• Adapted from WME02/01 June 2024 Q2• impulse, vectors
Question
A particle of mass $0.5$ kg moves with velocity $(3\mathbf{i} - 2\mathbf{j})$ m s $^{-1}$ . It receives an impulse $\mathbf{I}$ and immediately afterwards moves with velocity $(\mathbf{i} + 4\mathbf{j})$ m s $^{-1}$ . Find (a) $\mathbf{I}$ , (b) the magnitude of $\mathbf{I}$ .
Step-by-step solution
1. Step 1
  Impulse equals change of momentum:
  $\mathbf{I} = m\mathbf{v} - m\mathbf{u} = m(\mathbf{v} - \mathbf{u})$
2. Step 2
  Substitute:
  $\mathbf{I} = 0.5\left[(\mathbf{i} + 4\mathbf{j}) - (3\mathbf{i} - 2\mathbf{j})\right] = 0.5(-2\mathbf{i} + 6\mathbf{j}) = -\mathbf{i} + 3\mathbf{j}$
3. Step 3
  (b) Magnitude:
  $|\mathbf{I}| = \sqrt{(-1)^{2} + 3^{2}} = \sqrt{10} \approx 3.16\,\text{N s}$
Answer
(a) $\mathbf{I} = -\mathbf{i} + 3\mathbf{j}$ N s. (b) $|\mathbf{I}| = \sqrt{10} \approx 3.16$ N s.
Examiner tip
(a) M1 for $\mathbf{I} = m(\mathbf{v} - \mathbf{u})$ . A1 for vector. (b) M1 for $|\mathbf{I}|$ via Pythagoras. A1 for $\sqrt{10}$ . B1 for units N s (or equivalently kg m s $^{-1}$ ). Sign errors on $\mathbf{v} - \mathbf{u}$ are typical — write out both vectors fully before subtracting.
3Particle striking a wall (6 marks, M2)
Extended• Adapted from WME02/01 January 2023 Q7• wall collision, restitution
Question
A ball of mass $0.2$ kg strikes a smooth vertical wall horizontally at $15$ m s $^{-1}$ , perpendicular to the wall. The coefficient of restitution between ball and wall is $e = 0.7$ . Find (a) the speed of the ball immediately after the collision, (b) the magnitude of the impulse the wall exerts on the ball.
Step-by-step solution
1. Step 1
  Take the direction towards the wall as positive. Before: $u = 15$ . After: $v = -e \cdot u$ since restitution against an immovable surface gives the post-collision speed of magnitude $eu$ in the opposite direction.
2. Step 2
  (a) Speed after collision:
  $|v| = e u = 0.7 \cdot 15 = 10.5\,\text{m s}^{-1}$
3. Step 3
  (b) Impulse $= \Delta p = m v - m u$ :
  $I = 0.2 \cdot (-10.5) - 0.2 \cdot 15 = -2.1 - 3 = -5.1\,\text{N s}$
4. Step 4
  Magnitude $|I| = 5.1$ N s, directed away from the wall (the wall pushes the ball back).
Answer
(a) $10.5$ m s $^{-1}$ . (b) $|I| = 5.1$ N s.
Examiner tip
(a) M1 for restitution against a wall. A1 for $10.5$ m/s. (b) M1 for $I = m v - m u$ with correct signs. A1 for $|I| = 5.1$ N s. B1 for units. Common error: stating impulse = $0.2 \cdot 10.5 = 2.1$ N s (forgetting that the change of momentum is $m(v - u)$ , with $u$ and $v$ having opposite signs).
4Successive collisions between three particles (10 marks, M2)
Extended• Adapted from WME02/01 June 2023 Q8• successive collisions, momentum, restitution
Question
Three particles $A$ ( $1$ kg), $B$ ( $2$ kg) and $C$ ( $3$ kg) lie in this order on a smooth horizontal line. $A$ is projected toward $B$ at $10$ m s $^{-1}$ ; $B$ and $C$ are initially at rest. The coefficient of restitution between $A$ and $B$ is $e_1 = 0.5$ , and between $B$ and $C$ is $e_2 = 0.6$ . Find the final velocity of each particle.
Step-by-step solution
1. Step 1
  Collision 1: $A$ at $10$ hits $B$ at rest. Take right (the direction of $A$ 's motion) as positive.
2. Step 2
  Momentum: $1(10) + 2(0) = 1 v_A + 2 v_B$ , i.e. $v_A + 2 v_B = 10$ (i). Restitution: $v_B - v_A = 0.5(10 - 0) = 5$ (ii).
3. Step 3
  From (ii): $v_A = v_B - 5$ . Substitute: $(v_B - 5) + 2 v_B = 10$ , so $3 v_B = 15$ , $v_B = 5$ m/s. Then $v_A = 0$ .
4. Step 4
  Post-collision-1: $A$ at rest, $B$ moves at $5$ m/s toward $C$ . (No collision between $A$ and $B$ subsequently unless $B$ rebounds back.)
5. Step 5
  Collision 2: $B$ at $5$ hits $C$ at rest. Momentum: $2(5) + 3(0) = 2 v_B' + 3 v_C$ , i.e. $2 v_B' + 3 v_C = 10$ (iii). Restitution: $v_C - v_B' = 0.6(5 - 0) = 3$ (iv).
6. Step 6
  From (iv): $v_B' = v_C - 3$ . Substitute: $2(v_C - 3) + 3 v_C = 10$ , so $5 v_C = 16$ , $v_C = 3.2$ m/s. Then $v_B' = 0.2$ m/s.
7. Step 7
  Check: $B$ moves at $0.2$ m/s in the same direction as $C$ , but slower — no further collision between $B$ and $C$ . And $B$ is moving away from $A$ , who is at rest — no further $A$ - $B$ collision.
Answer
Final velocities: $A = 0$ , $B = 0.2$ m s $^{-1}$ , $C = 3.2$ m s $^{-1}$ (all in the direction of $A$ 's initial motion).
Examiner tip
Collision 1: M1 for momentum, M1 for restitution, M1 for solving, A2 for both velocities. Collision 2: M1 for momentum, M1 for restitution, M1 for solving, A1 for both velocities. A1 final check that no further collision occurs. The 'no further collision' check is mandatory — students who omit it lose the final mark, even with correct velocities.
5Coalescing particles (perfectly inelastic) — find combined velocity and KE loss (6 marks, M2)
Core• Adapted from WME02/01 January 2024 Q1(b)• inelastic, coalesce, energy loss
Question
A particle $P$ of mass $4$ kg moves at $5$ m s $^{-1}$ on a smooth horizontal surface and collides with a stationary particle $Q$ of mass $6$ kg. After the collision, $P$ and $Q$ coalesce (move together as one). Find (a) the speed immediately after the collision, (b) the kinetic energy lost.
Step-by-step solution
1. Step 1
  Coalescence is perfectly inelastic ( $e = 0$ ). Momentum conservation:
  $4(5) + 6(0) = (4 + 6) v \Rightarrow v = \dfrac{20}{10} = 2\,\text{m s}^{-1}$
2. Step 2
  KE before: $\tfrac{1}{2}(4)(5)^{2} + 0 = 50$ J. KE after: $\tfrac{1}{2}(10)(2)^{2} = 20$ J.
3. Step 3
  KE lost = $50 - 20 = 30$ J.
Answer
(a) $v = 2$ m s $^{-1}$ . (b) KE lost = $30$ J.
Examiner tip
(a) M1 for momentum equation with combined mass. A1 for $v = 2$ . (b) M1 for KE before AND after. A1 for $30$ J. Coalescence is the maximal-energy-loss collision for a given approach speed.

Key Formulae — Impulses and collisions

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

Impulse-momentum
$\mathbf{I} = \Delta\mathbf{p} = m\mathbf{v} - m\mathbf{u}$
$\mathbf{I}$
impulse (N s, vector)
$m$
mass (kg)
$\mathbf{u}, \mathbf{v}$
velocities before and after
When to use
Whenever a sudden change of velocity occurs (collision, jerk, kick). Equivalently $\mathbf{I} = \mathbf{F} t$ for a constant force acting for time $t$ . In the IAL formula booklet under 'Mechanics — Impulse and momentum'.
Example
Particle, $m = 0.5$ , $\mathbf{u} = 3\mathbf{i} - 2\mathbf{j}$ , $\mathbf{v} = \mathbf{i} + 4\mathbf{j}$ : $\mathbf{I} = -\mathbf{i} + 3\mathbf{j}$ N s.
Conservation of momentum
$m_A u_A + m_B u_B = m_A v_A + m_B v_B$
$m_A, m_B$
masses
$u_A, u_B$
velocities before collision (with sign)
$v_A, v_B$
velocities after collision (with sign)
When to use
Any collision where no external impulsive force acts (e.g. friction with the surface is negligible during the brief collision). In 2D: applies component-by-component. NOT explicitly in the formula booklet — derive from impulse-momentum.
Example
2 kg at $6$ hits 3 kg at $-4$ on a smooth line: $2(6) + 3(-4) = 2 v_A + 3 v_B = 0$ .
Newton's law of restitution
$v_B - v_A = e(u_A - u_B)$
$e$
coefficient of restitution, $0 \leq e \leq 1$
$u_A - u_B$
speed of approach (before)
$v_B - v_A$
speed of separation (after)
When to use
Pair the restitution equation with momentum conservation. $e = 0$ : perfectly inelastic (coalesce or stick). $e = 1$ : perfectly elastic (KE conserved). In the IAL formula booklet.
Example
$u_A = 6$ , $u_B = -4$ , $e = 0.5$ : $v_B - v_A = 0.5 \cdot 10 = 5$ .
Collision with a smooth fixed wall
$v_{\text{after}} = -e \cdot u_{\text{before}}$
$e$
coefficient of restitution
$u_{\text{before}}, v_{\text{after}}$
perpendicular component of velocity (signed)
When to use
When a particle strikes a wall perpendicularly. For oblique collisions, the parallel component is unchanged; only the perpendicular component flips sign and is multiplied by $e$ . NOT in the formula booklet — derive from restitution with wall velocity $= 0$ .
Example
Ball at $15$ m/s strikes wall, $e = 0.7$ : rebounds at $10.5$ m/s.
Kinetic energy lost in a collision
$\Delta\text{KE} = \text{KE}_{\text{before}} - \text{KE}_{\text{after}} = \tfrac{1}{2}m_A u_A^{2} + \tfrac{1}{2}m_B u_B^{2} - \tfrac{1}{2}m_A v_A^{2} - \tfrac{1}{2}m_B v_B^{2}$
$\Delta\text{KE}$
energy lost (always $\geq 0$ , with $= 0$ for elastic)
When to use
When asked for the kinetic energy lost or for the elasticity of the collision. NOT in the formula booklet — compute directly.
Example
Direct collision (worked example 1): $60 - 15 = 45$ J lost.

Key Definitions and Keywords — Impulses and collisions

Definitions to memorise and the exact keywords mark schemes credit for impulses and collisions answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Momentum
Examiner keyword
Vector quantity $\mathbf{p} = m\mathbf{v}$ . SI units: kg m s $^{-1}$ , equivalently N s. Sign matters in 1D problems — take a positive direction and apply it consistently.
Impulse
Examiner keyword
Change of momentum: $\mathbf{I} = \Delta\mathbf{p}$ . For a constant force over time, $\mathbf{I} = \mathbf{F} t$ . SI units: N s.
Coefficient of restitution
Examiner keyword
The ratio $e = \dfrac{\text{speed of separation}}{\text{speed of approach}}$ for a collision between two bodies. A property of the colliding pair, not of either body alone. Bounded: $0 \leq e \leq 1$ .
Elastic collision
Examiner keyword
A collision with $e = 1$ : kinetic energy is conserved (no KE lost). Idealisation — most collisions in M2 problems are partially elastic ( $0 < e < 1$ ).
Inelastic collision
Examiner keyword
Any collision with $e < 1$ — some KE is lost (typically as heat or sound). 'Perfectly inelastic' or 'coalesce' means $e = 0$ : the particles move together with a common velocity afterwards.
Speed of approach / separation
Examiner keyword
Speed of approach $= u_A - u_B$ when $A$ is moving toward $B$ (signs taken so this is positive). Speed of separation $= v_B - v_A$ after the collision. Both are scalars by construction. Newton's law of restitution: separation = $e \times$ approach.

Common Mistakes and Misconceptions — Impulses and collisions

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

✕Sign confusion in momentum and restitution equations
WME02/01 examiner reports — annual
Why it happens
Students forget to assign signs consistently.
How to avoid it
Declare positive direction at the start (typically the direction of one particle's initial motion). Apply signs consistently to ALL velocities (negative for opposite direction). After the algebra, check: positive result = same direction; negative = reversed.
✕Writing the restitution equation as $v_A - v_B = e(u_A - u_B)$
Why it happens
Memorisation slip — students miss that separation is $v_B - v_A$ , not $v_A - v_B$ , when $A$ was behind $B$ before.
How to avoid it
Restitution is ALWAYS $\dfrac{\text{separation}}{\text{approach}} = e$ with both quantities positive. Set it up using the physical interpretation: 'after the collision, the gap between them opens at speed $v_B - v_A$ '.
✕Using 'conservation of kinetic energy' instead of conservation of momentum
Why it happens
Energy conservation is the dominant theme in work-energy questions.
How to avoid it
KE is conserved ONLY for $e = 1$ (perfectly elastic). For any $e < 1$ , KE is lost. Momentum is ALWAYS conserved in a collision (provided no external impulses).
✕Computing wall impulse as $m v$ or $m u$ instead of $m(v - u)$
WME02/01 examiner reports — frequent
Why it happens
Treating impulse as 'final momentum' rather than 'change of momentum'.
How to avoid it
Impulse is ALWAYS $\Delta p = m(v - u)$ . For wall collisions, $v$ and $u$ have opposite signs, so $|I| = m(|u| + |v|) = m u (1 + e)$ for a perpendicular hit.
✕Not checking for further collisions in multi-particle problems
Why it happens
Once the algebra gives velocities, students move on.
How to avoid it
After each collision, check (i) whether the rear particle is now moving faster than the front (would catch up — another collision), (ii) whether any particle has rebounded backward toward an earlier particle (could collide again). Make the check explicit.
✕Omitting units on impulse (or quoting kg m/s instead of N s)
Why it happens
Both are correct SI but the convention is N s.
How to avoid it
Quote N s on impulse problems. They're dimensionally equivalent (1 N s = 1 kg m s $^{-1}$ ), but N s is the conventional unit at IAL.

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.

Impulses and collisions — frequently asked questions

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

Impulses and collisions

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Direct collision: find velocities and KE lost (8 marks, M2)

2Impulse on a particle in 2D (5 marks, M2)

3Particle striking a wall (6 marks, M2)

4Successive collisions between three particles (10 marks, M2)

5Coalescing particles (perfectly inelastic) — find combined velocity and KE loss (6 marks, M2)

Impulse-momentum

Conservation of momentum

Newton's law of restitution

Collision with a smooth fixed wall

Kinetic energy lost in a collision

Momentum

Impulse

Coefficient of restitution

Elastic collision

Inelastic collision

Speed of approach / separation

✕Sign confusion in momentum and restitution equations

✕Writing the restitution equation as vA−vB=e(uA−uB)v_A - v_B = e(u_A - u_B)vA​−vB​=e(uA​−uB​)

✕Using 'conservation of kinetic energy' instead of conservation of momentum

✕Computing wall impulse as mvm vmv or mum umu instead of m(v−u)m(v - u)m(v−u)

✕Not checking for further collisions in multi-particle problems

✕Omitting units on impulse (or quoting kg m/s instead of N s)

Practice questions

Past paper style quiz

Video lesson

Impulses and collisions — frequently asked questions

✕Writing the restitution equation as $v_A - v_B = e(u_A - u_B)$

✕Computing wall impulse as $m v$ or $m u$ instead of $m(v - u)$