# Momentum and Impulse

> CIE A-Level Further Mathematics · 9231 Further Mechanics
> Source: https://www.owlsprep.com/study/cie-9231-u3-momentum-and-impulse/

This sub-topic extends core momentum concepts, covering impulse for variable forces, two-dimensional collisions, and momentum conservation, with skills assessed frequently in CIE 9231 Further Mechanics papers.

**Prerequisites:** [Core A-Level linear momentum and impulse](https://www.owlsprep.com/study/cie-9709-mechanics-momentum-impulse/)

## Learning objectives

- Calculate linear momentum and impulse for constant and variable forces
- Apply the impulse-momentum principle to 1D and 2D motion
- Solve problems involving collisions between particles and fixed surfaces
- Use conservation of linear momentum for isolated systems

## Core Definitions of Momentum and Impulse

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

*Notation:* $\vec{p} = m\vec{v}$

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

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

*Notation:* $\vec{J} = \int_{t_1}^{t_2} \vec{F}(t) dt$

*Example:* A constant force of 10 N acting for 2 s has impulse $20 \text{ Ns}$ in the direction of the force.

**Worked example:** A variable force $F(t) = 2t + 4$ 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. $$J = mv - 0 \implies v = \frac{J}{m} = \frac{45}{3} = 15 \text{ m s}^{-1}$$

## The Impulse-Momentum Principle

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

*Starting from:* Newton's Second Law: $\vec{F} = \frac{d\vec{p}}{dt}$

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

*Conclusion:* Evaluating the integral gives $\vec{J} = \int_{t_1}^{t_2} \vec{F} dt = p_2 - p_1 = \Delta \vec{p}$, so impulse equals change in momentum.

**Worked example:** A 0.5 kg ball has initial velocity $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$ m s⁻¹. It is struck by a bat that gives it an impulse of $\begin{pmatrix} 2 \\ -5 \end{pmatrix}$ Ns. Find its final velocity.

1. Rearrange the impulse-momentum principle to solve for final velocity $\vec{v_2}$:
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. $$\vec{v_2} = \begin{pmatrix} 8 \\ -7 \end{pmatrix} \text{ m s}^{-1}$$

## Conservation of Linear Momentum

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).

> **tip**
>
> In 2D problems, resolve all momentum into perpendicular x and y components. Each component is conserved independently.

**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. $$4 = -2 + 3v_B \implies v_B = 2 \text{ m s}^{-1}$$

## Collisions with Fixed Smooth Surfaces

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 $10\sqrt{2}$ 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 $v_{parallel} = 10$ 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 $v_{perp}$:
6. $$14m = m v_{perp} + 10m \implies v_{perp} = 4 \text{ m s}^{-1}$$
7. Calculate the angle $\theta$ of rebound to the plane:
8. $$\tan\theta = \frac{v_{perp}}{v_{parallel}} = \frac{4}{10} = 0.4 \implies \theta \approx 21.8^\circ$$

## Common pitfalls

- **Wrong:** Adding magnitudes of momentum instead of vector components in 2D problems
  - Why it fails: Momentum is a vector, so direction must be accounted for via components
  - Correct: Resolve all momentum vectors into perpendicular components, add/subtract components, keep sign conventions consistent
- **Wrong:** Mixing up sign directions for velocity when calculating rebound impulse
  - Why it fails: Change in momentum is final minus initial, so incorrect signs give wrong impulse magnitude
  - Correct: Define a fixed positive direction before starting, substitute velocities with their correct signs when calculating $\Delta p$
- **Wrong:** Applying conservation of momentum when there is an external resultant force
  - Why it fails: Momentum conservation only holds for isolated systems with no net external force
  - Correct: Check for external forces (e.g. weight, impulse from fixed surfaces) before using momentum conservation
- **Wrong:** Treating impulse as a scalar quantity, ignoring direction
  - Why it fails: Impulse follows the same vector rules as momentum, direction is critical for 2D problems
  - Correct: Always represent impulse as a vector, resolve into components for 2D calculations

## Cheatsheet

| Concept | Formula | Key Notes |
| --- | --- | --- |
| Linear Momentum | $\vec{p} = m\vec{v}$ | Vector, units: kg m s⁻¹ = Ns |
| Impulse (Constant F) | $\vec{J} = \vec{F} \Delta t$ | Vector, magnitude = force × time |
| Impulse (Variable F) | $\vec{J} = \int_{t_1}^{t_2} \vec{F}(t) dt$ | Equals area under F-t graph |
| Impulse-Momentum Principle | $\vec{J} = \Delta \vec{p} = \vec{p}_2 - \vec{p}_1$ | Derived from Newton's Second Law |
| Conservation of Momentum | $\sum \vec{p}_{init} = \sum \vec{p}_{final}$ | Holds if no external resultant force |
| Smooth Fixed Surface Collision | Parallel momentum conserved | Impulse only acts perpendicular to surface |

## 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.

- [Work and Energy](https://www.owlsprep.com/study/cie-9231-u3-work-and-energy/)
- [Circular Motion](https://www.owlsprep.com/study/cie-9231-u3-circular-motion/)
- [Equilibrium of rigid bodies](https://www.owlsprep.com/study/cie-9231-u3-equilibrium-of-rigid-bodies/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/cie-9231-u3-momentum-and-impulse/