# Dynamics of a particle moving in a straight line or plane

> Edexcel International A-Level Mathematics · IAL M1
> Source: https://www.owlsprep.com/study/edexcel-ial-math-m1-dynamics-of-a-particle-moving/

This guide covers all core M1 dynamics content for Edexcel IAL, including Newton's laws, connected particles, friction, impulse, and momentum, with exam-standard worked examples and common mistake avoidance tips.

**Prerequisites:** [Constant acceleration (suvat) equations](https://www.owlsprep.com/study/edexcel-ial-math-m1-constant-acceleration/); [2D vector arithmetic](https://www.owlsprep.com/study/edexcel-ial-math-m1-vectors/)

## Learning objectives

- Apply Newton's laws of motion to particles with constant acceleration in scalar or vector form
- Solve connected particle problems including pulleys, inclined planes, and post-impact force changes
- Use the impulse-momentum principle and conservation of momentum for 1D collisions
- Calculate kinetic friction using F=μR for moving particles

## Newton's Laws of Motion for Constant Acceleration

**Newton's Second Law of Motion** — The resultant force acting on a particle equals the product of its mass and acceleration: \(F_{\text{net}} = ma\). For 2D motion, this applies separately to i and j components.

*Example:* A 2kg particle with resultant force (6i - 4j) N has acceleration (3i - 2j) m s⁻².

**Worked example:** A 3kg particle is acted on by two forces: \(F_1 = (2i + 5j)\) N and \(F_2 = (7i - 11j)\) N. Calculate the magnitude of the particle's acceleration, to 2 decimal places.

1. Step 1: Calculate resultant force by adding vector components

   $$F_{\text{net}} = (2+7)i + (5-11)j = 9i -6j \text{ N}$$
2. Step 2: Apply \(F_{net}=ma\) to find acceleration vector

   $$a = \frac{F_{net}}{m} = \frac{9i -6j}{3} = 3i - 2j \text{ m s}^{-2}$$
3. Step 3: Calculate magnitude using Pythagoras

   $$|a| = \sqrt{3^2 + (-2)^2} = \sqrt{13} \approx 3.61 \text{ m s}^{-2}$$

> **Exam tip:** Always resolve forces into perpendicular components (horizontal/vertical or parallel/perpendicular to an incline) before applying Newton's second law.

*Calculator:* allowed

## Connected Particles and Inclined Planes

Connected particle problems involve two or more masses linked by light inextensible strings, often over smooth fixed pulleys. For rough inclined planes, include kinetic friction if the particle is moving.

**Light inextensible string** — String with negligible mass that does not stretch, meaning tension is equal throughout the string, and connected particles accelerate at the same magnitude.

*Example:* Two masses hanging over a pulley accelerate at the same rate, with equal tension on both sides of the pulley.

**Worked example:** A 2kg block rests on a rough horizontal table, connected by a light inextensible string over a smooth fixed pulley to a 3kg mass hanging vertically. The coefficient of kinetic friction between the block and table is 0.4. Calculate the acceleration of the system, to 2 decimal places.

1. Step 1: Assign positive direction as the hanging mass moving downward, so both masses accelerate at rate \(a\)
2. Step 2: Write force equation for hanging mass

   $$3g - T = 3a$$
3. Step 3: Calculate normal reaction for block: \(R = 2g\), so friction \(F = \mu R = 0.8g\). Write force equation for block:

   $$T - 0.8g = 2a$$
4. Step 4: Add equations to eliminate T

   $$3g - 0.8g = 5a \implies 2.2g = 5a$$
5. Step 5: Solve for a

   $$a = \frac{2.2*9.8}{5} = 4.31 \text{ m s}^{-2}$$

> **Exam tip:** If a connected particle hits the ground, the string goes slack so tension becomes zero, and the remaining particle moves under new forces (e.g. friction) with a new acceleration.

*Calculator:* allowed

## Impulse and 1D Conservation of Momentum

**Impulse-Momentum Principle** — Impulse equals the change in momentum of a particle: \(I = mv - mu\), where \(u\) is initial velocity, \(v\) is final velocity, and \(m\) is mass. Momentum is conserved for 1D collisions with no external forces: total momentum before = total momentum after.

**Worked example:** A 0.5kg ball moving at 4 m s⁻¹ right collides directly with a 0.3kg ball moving at 2 m s⁻¹ left. After collision, the 0.3kg ball moves at 3 m s⁻¹ right. Calculate the final velocity of the 0.5kg ball, stating direction.

1. Step 1: Assign positive direction as right, so \(u_1 = 4\) m s⁻¹, \(u_2 = -2\) m s⁻¹, \(v_2 = 3\) m s⁻¹
2. Step 2: Apply conservation of momentum

   $$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$
3. Step 3: Substitute values

   $$(0.5*4) + (0.3*(-2)) = 0.5v_1 + (0.3*3)$$
4. $$2 - 0.6 = 0.5v_1 + 0.9 \implies 0.5v_1 = 0.5$$
5. Step 4: Final result: \(v_1 = 1\) m s⁻¹ to the right.

> **Exam tip:** Always state your chosen positive direction for velocity in momentum problems to avoid sign errors and show clear reasoning to examiners.

*Calculator:* allowed

## Kinetic Friction Calculations

**Kinetic Friction** — Friction force acting on a moving particle, given by \(F = \mu R\), where \(\mu\) is the coefficient of kinetic friction and \(R\) is the normal reaction force. Friction always acts opposite to the direction of motion.

**Worked example:** A 5kg particle slides down a rough plane inclined at 30° to the horizontal. The coefficient of kinetic friction is 0.2. Calculate the acceleration of the particle, to 2 decimal places.

1. Step 1: Resolve perpendicular to plane to find normal reaction

   $$R = 5g\cos(30^\circ)$$
2. Step 2: Calculate friction force

   $$F = \mu R = 0.2*5g\cos(30^\circ) = g\cos(30^\circ)$$
3. Step 3: Resolve parallel to plane, apply \(F_{net}=ma\)

   $$5g\sin(30^\circ) - F = 5a$$
4. Step 4: Solve for a

   $$24.5 - 8.487 = 5a \implies a \approx 3.20 \text{ m s}^{-2}$$

> **Exam tip:** Use \(\sin(\theta)\) for the parallel component of weight on an incline, and \(\cos(\theta)\) for the perpendicular component. Mixing these is a very common error.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using F=μR for stationary particles
  - Why it fails: F=μR only applies to kinetic friction for moving particles. Static friction is not covered in M1 and this relationship does not hold for stationary objects.
  - Correct: Only use F=μR when the question explicitly states the particle is moving, or you have confirmed it is in motion.
- **Wrong:** Ignoring velocity direction in momentum calculations
  - Why it fails: Momentum is a vector quantity, so particles moving opposite to your chosen positive direction have negative velocity values. Omitting signs leads to incorrect total momentum sums.
  - Correct: Define a positive direction at the start of all momentum problems, and assign negative values to velocities opposite to this direction.
- **Wrong:** Using g=10 m s⁻² instead of 9.8 m s⁻²
  - Why it fails: Edexcel IAL M1 explicitly requires g=9.8 m s⁻². Using g=10 will lead to incorrect answers and lose accuracy marks.
  - Correct: Always use g=9.8 m s⁻² in all calculations unless the question explicitly states otherwise.
- **Wrong:** Assuming tension differs on either side of a smooth pulley
  - Why it fails: Smooth fixed pulleys have no friction, so tension is identical throughout a light inextensible string passing over them.
  - Correct: Use the same tension value for both sides of a smooth fixed pulley when writing force equations for connected particles.
- **Wrong:** Applying conservation of momentum when external forces act
  - Why it fails: Conservation of momentum only holds if there are no external forces (e.g. friction, applied forces) acting on the system of colliding particles.
  - Correct: Only use conservation of momentum for direct 1D collisions between two particles with no external forces acting during impact.

## Cheatsheet

| Concept | Formula/Rule | Key Note |
| --- | --- | --- |
| Newton's Second Law (scalar) | \(F_{\text{net}} = ma\) | Apply to resolved force components |
| Newton's Second Law (vector) | \(\mathbf{F}_{\text{net}} = m\mathbf{a}\) | Apply separately to i and j components |
| Kinetic Friction | \(F = \mu R\) | Only use for moving particles |
| Impulse | \(I = mv - mu\) | Vector quantity, sign depends on direction |
| Conservation of Momentum | \(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\) | 1D only, no external forces, assign positive direction first |
| Connected Particles (smooth pulley) | Tension equal on both sides, acceleration magnitude identical | If one particle hits ground, tension becomes zero |

## What's next

Now that you have mastered core M1 dynamics concepts, you are ready to practice full exam-style questions and move on to subsequent M1 topics. Dynamics accounts for ~25% of marks in most M1 papers, so practicing mixed problem sets will help you build speed and accuracy for your exam. Make sure you also review vector applications for 2D force problems, as these are often combined with dynamics in longer 6-8 mark exam questions. Once you are comfortable with dynamics, you can move on to the next M1 topic: moments.

---

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-math-m1-dynamics-of-a-particle-moving/