Momentum and Impulse

CIE A-Level Further Mathematics· Unit 3: Further Mechanics, Topic 1· 20 min read

1. Core Definitions of Momentum and Impulse★★☆☆☆⏱ 5 min

📘 Definition

Linear Momentum

$p = m v$

A vector quantity equal to the product of a particle's mass and velocity, with direction matching the velocity of the particle.

Example:

A 2 kg mass moving at $3 m s^{- 1}$ right has momentum $6 kg m s^{- 1}$ right.

📘 Definition

Impulse

$J = \int_{t_{1}}^{t_{2}} F (t) d t$

The integral of force over the time interval it acts, a vector quantity equal to the total change in momentum of a particle.

Example:

A constant force of 10 N acting for 2 s has impulse $20 Ns$ in the direction of the force.

📐 Worked Example

A variable force $F (t) = 2 t + 4$ N acts on a 3 kg particle initially at rest along a straight line. Find the final speed after 5 seconds.

1
Calculate total impulse by integrating the force from $t = 0$ to $t = 5$ :
2
$J = \int_0^5 (2t + 4) dt = \left[t^2 + 4t\right]_0^5 = 25 + 20 = 45 \text{ Ns}$
3
Apply the impulse-momentum principle, where initial momentum is 0:
4
$J = mv - 0 \implies v = \frac{J}{m} = \frac{45}{3} = 15 \text{ m s}^{-1}$

2. The Impulse-Momentum Principle★★★☆☆⏱ 7 min

The impulse-momentum principle is derived directly from Newton's Second Law of Motion, and works for both constant and variable forces, and for vector motion in multiple dimensions.

🔬 Derivation

Goal:

Derive the impulse-momentum principle from Newton's Second Law

Starting from:

Newton's Second Law: $F = \frac{d p}{d t}$

1
Rearrange and integrate both sides over the time interval $t_{1}$ (initial) to $t_{2}$ (final):
2
$\int_{t_1}^{t_2} \vec{F} dt = \int_{p_1}^{p_2} d\vec{p}$

Result:

Evaluating the integral gives $J = \int_{t_{1}}^{t_{2}} F d t = p_{2} - p_{1} = Δ p$ , so impulse equals change in momentum.

📐 Worked Example

A 0.5 kg ball has initial velocity $(43)$ m s⁻¹. It is struck by a bat that gives it an impulse of $(2 - 5)$ Ns. Find its final velocity.

1
Rearrange the impulse-momentum principle to solve for final velocity $v_{2}$ :
2
$\vec{J} = m(\vec{v_2} - \vec{v_1}) \implies \vec{v_2} = \vec{v_1} + \frac{\vec{J}}{m}$
3
Substitute the given values for mass, initial velocity and impulse:
4
$\vec{v_2} = \begin{pmatrix} 4 \\ 3 \end{pmatrix} + \frac{1}{0.5}\begin{pmatrix} 2 \\ -5 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix} + \begin{pmatrix} 4 \\ -10 \end{pmatrix}$
5
Add the vectors to get the final result:
6
$\vec{v_2} = \begin{pmatrix} 8 \\ -7 \end{pmatrix} \text{ m s}^{-1}$

3. Conservation of Linear Momentum★★☆☆☆⏱ 6 min

For any isolated system of particles with no external resultant force acting, the total linear momentum of the system is constant. Internal forces between particles do not change the total momentum, because they come in equal and opposite pairs (Newton's Third Law).

📐 Worked Example

Two particles A (mass 2 kg) and B (mass 3 kg) move towards each other along a straight line. A has initial speed 5 m s⁻¹, B has initial speed 2 m s⁻¹. After collision, A moves opposite its original direction at 1 m s⁻¹. Find B's final speed.

1
Take A's original direction as positive, write total initial momentum:
2
$p_{init} = m_A u_A + m_B u_B = (2)(5) + (3)(-2) = 10 - 6 = 4 \text{ kg m s}^{-1}$
3
Write total final momentum, with A's final velocity = -1 m s⁻¹:
4
$p_{final} = (2)(-1) + 3v_B = -2 + 3v_B$
5
Equate initial and final momentum (no external force):
6
$4 = -2 + 3v_B \implies v_B = 2 \text{ m s}^{-1}$

4. Collisions with Fixed Smooth Surfaces★★★★☆⏱ 8 min

When a smooth particle collides with a fixed surface, impulse only acts perpendicular to the surface, as there is no friction force parallel to the surface. This means the parallel component of the particle's momentum (and velocity) does not change during collision.

📐 Worked Example

A smooth ball of mass m hits a horizontal plane at 45° with speed $102$ m s⁻¹. The impulse from the plane is 14m Ns. Find the angle the rebound velocity makes with the plane.

1
Resolve initial velocity into parallel (to plane) and perpendicular (towards plane = positive) components:
2
$u_{parallel} = 10\sqrt{2} \cos 45^\circ = 10 \text{ m s}^{-1}, \quad u_{perpendicular} = 10\sqrt{2} \sin 45^\circ = 10 \text{ m s}^{-1}$
3
Parallel component is unchanged, so $v_{p a r a l l e l} = 10$ m s⁻¹. Use impulse = change in perpendicular momentum (final velocity is away from plane, so negative):
4
$J = (-mv_{perp}) - (-m u_{perp})^{-1} = m v_{perp} + 10m$
5
Substitute J = 14m and solve for $v_{p er p}$ :
6
$14m = m v_{perp} + 10m \implies v_{perp} = 4 \text{ m s}^{-1}$
7
Calculate the angle $θ$ of rebound to the plane:
8
$\tan\theta = \frac{v_{perp}}{v_{parallel}} = \frac{4}{10} = 0.4 \implies \theta \approx 21.8^\circ$

5. Common Pitfalls

Wrong move:

Adding magnitudes of momentum instead of vector components in 2D problems

Why:

Momentum is a vector, so direction must be accounted for via components

Correct move:

Resolve all momentum vectors into perpendicular components, add/subtract components, keep sign conventions consistent

Wrong move:

Mixing up sign directions for velocity when calculating rebound impulse

Why:

Change in momentum is final minus initial, so incorrect signs give wrong impulse magnitude

Correct move:

Define a fixed positive direction before starting, substitute velocities with their correct signs when calculating $Δ p$

Wrong move:

Applying conservation of momentum when there is an external resultant force

Why:

Momentum conservation only holds for isolated systems with no net external force

Correct move:

Check for external forces (e.g. weight, impulse from fixed surfaces) before using momentum conservation

Wrong move:

Treating impulse as a scalar quantity, ignoring direction

Why:

Impulse follows the same vector rules as momentum, direction is critical for 2D problems

Correct move:

Always represent impulse as a vector, resolve into components for 2D calculations

6. Quick Reference Cheatsheet

Concept	Formula	Key Notes
Linear Momentum	$p = m v$	Vector, units: kg m s⁻¹ = Ns
Impulse (Constant F)	$J = F Δ t$	Vector, magnitude = force × time
Impulse (Variable F)	$J = \int_{t_{1}}^{t_{2}} F (t) d t$	Equals area under F-t graph
Impulse-Momentum Principle	$J = Δ p = p_{2} - p_{1}$	Derived from Newton's Second Law
Conservation of Momentum	$\sum p_{ini t} = \sum p_{f ina l}$	Holds if no external resultant force
Smooth Fixed Surface Collision	Parallel momentum conserved	Impulse only acts perpendicular to surface

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Variable force impulse calculation
2021 · 2
2D impulse velocity problem
2023 · 1
Collision with fixed surface impulse

Going deeper

syllabusCIE 9231 Further Maths SyllabusOfficial unit 3 topic reference

What's Next

Momentum and impulse form the foundation for all collision and variable force problems in CIE 9231 Further Mechanics. This sub-topic extends directly into oblique collisions between two moving particles, where you will combine momentum conservation with the coefficient of restitution to solve 2D collision problems. Mastery of vector momentum and impulse is also required for topics like variable force work-energy, connected particle motion, and impacts in circular motion. Consistent application of sign conventions and vector component resolution will help you score full marks on these high-weight exam questions.