# Further Mechanics (Edexcel IAL Physics Unit 4)

> Physics · Edexcel IAL
> Source: https://www.owlsprep.com/study/edexcel-ial-physics-u4-further-mechanics/

This guide covers all Edexcel IAL Physics Unit 4 Further Mechanics content, including impulse, 2D momentum conservation, circular motion kinematics/dynamics, and required core practicals 9 and 10, aligned with exam rules.

**Prerequisites:** [Edexcel IAL Physics Unit 1 linear momentum basics](https://www.owlsprep.com/study/edexcel-ial-physics-u1-motion-forces-energy/); [Vector component resolution skills](https://www.owlsprep.com/study/edexcel-ial-physics-u1-maths-skills/)

## Learning objectives

- Calculate impulse using $F\Delta t = \Delta p$ and apply to collision scenarios
- Solve 2D conservation of momentum problems and distinguish elastic vs inelastic collisions
- Derive and use $E_k = p^2/2m$ for non-relativistic particles
- Convert between angular displacement units, use $v=\omega r$ and $T=2\pi/\omega$
- Derive centripetal acceleration via vector diagrams and apply centripetal force equations
- Describe and interpret results for Core Practicals 9 and 10

## Impulse and Momentum Relationships

**Impulse** — Impulse is the change in momentum of an object, equal to the product of the average force applied to the object and the time interval the force acts for. The area under a force-time graph equals total impulse.

*Notation:* $J = F\Delta t = \Delta p = m(v-u)$

Impulse is derived directly from Newton's second law of motion, rearranged without calculus. You are also required to derive the kinetic energy formula $E_k = \frac{p^2}{2m}$ algebraically for exam questions.

**Derivation:** Derive $E_k = \frac{p^2}{2m}$

1. Start with the standard non-relativistic kinetic energy formula:
2. $$E_k = \frac{1}{2}mv^2$$
3. Recall linear momentum $p = mv$, so rearrange to get $v = \frac{p}{m}$
4. Substitute $v = \frac{p}{m}$ into the kinetic energy equation:
5. $$E_k = \frac{1}{2}m\left(\frac{p}{m}\right)^2 = \frac{p^2}{2m}$$

*Conclusion:* This formula lets you relate kinetic energy directly to momentum without calculating velocity first.

**Worked example:** A 0.15 kg ball travelling horizontally at 12 m s⁻¹ is hit by a bat and rebounds horizontally at 18 m s⁻¹. The bat is in contact with the ball for 0.002 s. Calculate the average force exerted by the bat on the ball.

1. Step 1: Define the initial direction of the ball as positive, so final velocity is negative
2. Step 2: Calculate change in momentum:
3. $$\Delta p = m(v - u) = 0.15(-18 - 12) = -4.5 \text{ kg m s}^{-1}$$
4. Step 3: Rearrange the impulse formula to solve for force:
5. $$F = \frac{\Delta p}{\Delta t} = \frac{-4.5}{0.002} = -2250 \text{ N}$$
6. The negative sign indicates force acts opposite to the initial direction, so magnitude is 2250 N.

> **Exam tip:** Always explicitly define your positive direction when calculating change in momentum to avoid sign errors, especially for rebound scenarios. Marks are awarded for correct sign convention as well as numerical answer.

## 2D Conservation of Momentum and Collisions

For any collision with no external resultant force, total linear momentum is conserved in all directions. For 2D collisions, resolve velocity vectors into horizontal (x) and vertical (y) components, apply conservation of momentum separately to each axis, then combine components to find final velocity magnitude and direction.

**Elastic vs Inelastic Collisions** — Elastic collisions conserve both total momentum and total kinetic energy. Inelastic collisions conserve total momentum but not total kinetic energy (energy is lost as heat, sound or deformation).

*Example:* A billiard ball collision is nearly elastic; a car collision is highly inelastic.

**Worked example:** A 2 kg snooker ball travelling at 3 m s⁻¹ along the x-axis collides with a stationary 2 kg snooker ball. After collision, the first ball moves at 1.5 m s⁻¹ at 60° above the x-axis. Calculate the velocity (magnitude and direction) of the second ball, and state if the collision is elastic.

1. Step 1: Apply conservation of momentum in the x-direction:
2. $$(2 \times 3) + 0 = (2 \times 1.5\cos60°) + (2 \times v_{2x}) \implies v_{2x} = 2.25 \text{ m s}^{-1}$$
3. Step 2: Apply conservation of momentum in the y-direction (initial total y momentum = 0):
4. $$0 = (2 \times 1.5\sin60°) + (2 \times v_{2y}) \implies v_{2y} = -1.30 \text{ m s}^{-1} (below x-axis)$$
5. Step 3: Calculate magnitude of second ball's velocity:
6. $$v_2 = \sqrt{2.25^2 + 1.30^2} = 2.60 \text{ m s}^{-1}$$
7. Step 4: Calculate direction: $\theta = \arctan(|1.30/2.25|) = 30°$ below the x-axis
8. Step 5: Check KE conservation: Initial $E_k = 9$ J, Final $E_k = 2.25 + 6.76 = 9.01$ J (rounding error), so collision is elastic.

> **info**
>
> Core Practical 10 uses ICT (motion sensors or video analysis) to record velocities of colliding trolleys/pucks in 2D to verify conservation of momentum. You should be able to describe equipment setup, error reduction, and result interpretation.

> **Exam tip:** Never apply conservation of momentum to resultant velocity directly without resolving into components. Always state the direction of final velocity with reference to a given axis (e.g. 30° below the horizontal).

## Circular Motion Kinematics

Circular motion describes an object moving at constant speed around a fixed circular path. Angular displacement is the angle rotated by the object around the centre, measured in degrees or radians. Convert between units using $1 \text{ rad} = \frac{180°}{\pi}$.

**Angular Velocity** — Angular velocity is the rate of change of angular displacement, measured in radians per second (rad s⁻¹). $T$ is the time period for one full rotation, $f$ is frequency of rotation, $r$ is path radius, $v$ is linear speed.

*Notation:* $\omega = \frac{\Delta \theta}{\Delta t} = \frac{2\pi}{T} = 2\pi f$, $v = \omega r$

**Worked example:** A car travels around a circular roundabout of radius 25 m at a constant linear speed of 12 m s⁻¹. Calculate its angular velocity and time taken to complete one full lap.

1. Step 1: Use $v = \omega r$ to calculate angular velocity:
2. $$\omega = \frac{v}{r} = \frac{12}{25} = 0.48 \text{ rad s}^{-1}$$
3. Step 2: Use $T = \frac{2\pi}{\omega}$ to calculate time period:
4. $$T = \frac{2\pi}{0.48} = 13.1 \text{ s (3 s.f.)}$$

> **Exam tip:** Always convert angular displacement to radians before calculating angular velocity or using other circular motion formulae. Answers in degrees per second will not be awarded marks unless specifically requested.

## Centripetal Acceleration and Force

Even if an object moves at constant speed in a circular path, it accelerates because its direction (and hence velocity vector) constantly changes. This acceleration is directed towards the centre of the circle, called centripetal acceleration. You must derive this using vector diagrams, not calculus.

**Derivation:** Derive centripetal acceleration $a = \frac{v^2}{r}$ using vector change in velocity

1. Consider an object moving at constant speed $v$ around a circle of radius $r$, moving small angular displacement $\Delta \theta$ in time $\Delta t$.
2. Initial velocity $\vec{v_1}$ and final velocity $\vec{v_2}$ have equal magnitude $v$, separated by angle $\Delta \theta$.
3. Draw a vector triangle for $\Delta \vec{v} = \vec{v_2} - \vec{v_1}$. For small $\Delta \theta$, magnitude of $\Delta v \approx v\Delta \theta$.
4. Acceleration $a = \frac{\Delta v}{\Delta t} = \frac{v\Delta \theta}{\Delta t}$.
5. Substitute $\omega = \frac{\Delta \theta}{\Delta t}$ and $\omega = \frac{v}{r}$:
6. $$a = v\omega = \frac{v^2}{r} = r\omega^2$$

*Conclusion:* Centripetal acceleration is always directed towards the centre of the circular path, perpendicular to instantaneous velocity.

A resultant force is required to produce centripetal acceleration, called centripetal force. This force also acts towards the centre of the circle, and can be provided by friction, tension, normal contact force, or other forces depending on the scenario. Centripetal force formulae: $F = \frac{mv^2}{r} = mr\omega^2$.

**Worked example:** A 1200 kg car travels around a flat circular bend of radius 50 m at 15 m s⁻¹. Calculate the minimum coefficient of friction between tyres and road required to prevent skidding.

1. Step 1: Centripetal force is provided by friction between tyres and road:
2. $$F_f = F_c = \frac{mv^2}{r}$$
3. Step 2: Maximum friction force $F_f = \mu R$, where normal contact force $R = mg$ on a flat road
4. Step 3: Equate and cancel mass $m$ from both sides:
5. $$\mu = \frac{v^2}{rg} = \frac{15^2}{50 \times 9.81} = 0.459 (3 s.f.)$$

> **info**
>
> Core Practical 9 investigates the relationship between force and change in momentum, or centripetal force for a mass moving in a horizontal circle. You should be able to identify variables and explain sources of experimental error.

> **Exam tip:** Centripetal force is not a new type of force: it is the name given to the resultant force that causes circular motion. Always state what force provides the centripetal force in a given scenario (e.g. friction, tension) to gain full marks.

## Common pitfalls

- **Wrong:** Using the same sign for initial and final velocity in rebound impulse calculations
  - Why it fails: Velocity is a vector, so direction changes require a sign change if you defined a positive axis
  - Correct: Explicitly define a positive direction at the start of any momentum calculation, and apply signs consistently to velocity values
- **Wrong:** Applying conservation of momentum to resultant velocity in 2D collisions instead of resolving into components
  - Why it fails: Momentum is conserved separately in each perpendicular direction, not in the resultant direction directly
  - Correct: Resolve all velocity vectors into x and y components, apply conservation of momentum to each axis separately, then combine components if needed
- **Wrong:** Using degrees instead of radians in circular motion formulae involving $\omega$
  - Why it fails: All standard circular motion formulae require $\omega$ to be in radians per second, not degrees per second
  - Correct: Convert all angular displacement values to radians before calculating angular velocity or using circular motion formulae
- **Wrong:** Treating centripetal force as an additional separate force on free-body diagrams
  - Why it fails: Centripetal force is the resultant of real forces (friction, tension, normal force) acting on the object, not a new force itself
  - Correct: Draw all real forces acting on the object, then state that the resultant of these forces provides the required centripetal force
- **Wrong:** Assuming all collisions are elastic unless stated otherwise
  - Why it fails: Most real collisions are inelastic, with some kinetic energy lost as heat, sound or deformation
  - Correct: Always calculate total initial and final kinetic energy to confirm if a collision is elastic, rather than assuming it is

## Cheatsheet

| Concept | Formulae | Key Exam Tip |
| --- | --- | --- |
| Impulse | $F\Delta t = \Delta p = m(v-u)$ | Define positive direction to avoid sign errors |
| KE-Momentum Relation | $E_k = \frac{p^2}{2m}$ | Use to link KE and momentum without velocity |
| 2D Momentum | Conserved in x and y axes separately | Resolve vectors into components first |
| Angular Kinematics | $\omega = \frac{2\pi}{T}$, $v = \omega r$ | Use radians only for $\omega$ |
| Centripetal Acceleration | $a = \frac{v^2}{r} = r\omega^2$ | Derive via vector triangles, not calculus |
| Centripetal Force | $F = \frac{mv^2}{r} = mr\omega^2$ | State what force provides centripetal force |

## What's next

Now that you have mastered Further Mechanics for Edexcel IAL Physics Unit 4, you are ready to move on to Electric and Magnetic Fields, where you will apply your force and motion knowledge to charged particles in fields. Practice past paper questions focused on this topic to reinforce your understanding, and review Core Practicals 9 and 10 in detail, as they often appear in 3-6 mark extended response questions. Your circular motion knowledge will also be built on later in Unit 5 when studying orbital motion under gravitational fields.

---

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/edexcel-ial-physics-u4-further-mechanics/