# Forces and momentum

> IB Physics SL · IB Physics 2025 SL
> Source: https://www.owlsprep.com/study/ib-physics-sl-u1-forces-and-momentum/

This subtopic covers core concepts of forces, Newton's laws of motion, momentum, and impulse for IB Physics SL. You will learn to analyze force systems and apply conservation rules to solve common exam problems.

**Prerequisites:** [Kinematics and linear motion equations](https://www.owlsprep.com/study/ib-physics-sl-u1-kinematics/)

## Learning objectives

- Draw accurate free-body diagrams for different force systems
- Apply Newton's three laws of motion to solve linear motion problems
- Resolve forces into perpendicular components for angled systems
- Use conservation of momentum to solve collision problems

## Force Types and Free-Body Diagrams

**Force** — A vector quantity describing the push or pull interaction on an object, measured in newtons (N).

*Notation:* \vec{F}

*Example:* The weight of a 10 kg mass is 98 N, directed vertically downwards.

Forces are split into two categories: contact forces (friction, tension, normal reaction) and non-contact forces (weight, gravity, electrostatic force). The first step for any force problem is drawing a free-body diagram (FBD) that shows only forces acting on your chosen object.

**Worked example:** Draw the free-body diagram for a 5 kg block sliding at constant speed down a rough 30° inclined plane.

1. 1. Isolate the block as your system, only include forces acting on the block.
2. 2. Add the weight force (always present for masses near Earth):
3. $$W = mg = 5 \times 9.8 = 49 \text{ N, vertically downward}$$
4. 3. Add the normal reaction force, acting perpendicular to the incline from the surface onto the block.
5. 4. Add kinetic friction, acting parallel to the incline opposite the direction of motion (up the incline).

> **Exam tip:** Always draw force arrows starting from the center of mass of the object to avoid losing marks in IB exams.

## Newton's Laws of Motion

**Newton's First Law (Inertia)** — A body at rest stays at rest, and a body at constant velocity stays at constant velocity, unless acted upon by a net external force.

Newton's second law relates net force to acceleration: $\vec{F}_{net} = m\vec{a}$, where $F_{net}$ is the vector sum of all forces, $m$ is mass, and $a$ is acceleration. Newton's third law states that every action force has an equal and opposite reaction force, which acts on a different body.

> **warning**
>
> Newton's third law action-reaction pairs never act on the same body, so they cannot cancel each other out.

**Worked example:** A 10 kg box is pulled horizontally along a frictionless surface by a rope with 50 N tension at 30° above the horizontal. Calculate the box's acceleration.

1. 1. Resolve tension into horizontal and vertical components:
2. $$T_x = T \cos\theta = 50 \cos 30^\circ \approx 43.3 \text{ N}$$
3. 2. There is no friction, so net horizontal force equals $T_x$.
4. 3. Apply Newton's second law $F_{net} = ma$:
5. $$a = \frac{F_{net}}{m} = \frac{43.3}{10} \approx 4.3 \text{ m s}^{-2}$$

## Force Resolution and Equilibrium

**Translational Equilibrium** — A system is in equilibrium when net external force is zero, so acceleration equals zero.

For forces acting at angles, resolve all vectors into perpendicular x and y components, then sum the components to find net force. For equilibrium, $\sum F_x = 0$ and $\sum F_y = 0$.

**Worked example:** A 2.0 kg picture is hung from a nail by two equal-length strings, each at 20° to the horizontal. Find the tension in each string.

1. 1. Let $T$ = tension in each string. Horizontal components are $T\cos20^\circ$ left and right, so they cancel: $\sum F_x = 0$.
2. 2. Sum vertical components: upward tension forces balance downward weight:
3. $$2T \sin 20^\circ = mg$$
4. 3. Rearrange to solve for T:
5. $$T = \frac{mg}{2 \sin 20^\circ} = \frac{2.0 \times 9.8}{2 \times 0.3420} \approx 29 \text{ N}$$

> **Exam tip:** Always label your positive direction when resolving forces to avoid sign errors.

## Momentum, Impulse and Conservation of Momentum

**Linear Momentum** — A vector quantity equal to the product of an object's mass and velocity: $p = mv$, units of kg m s⁻¹.

*Notation:* p

Impulse is the change in momentum of an object, equal to the product of average force and contact time: $J = F_{avg} \Delta t = \Delta p$. For an isolated system with no net external force, total momentum is conserved: total momentum before a collision equals total momentum after the collision.

**Worked example:** A 0.5 kg ball moving at 4 m s⁻¹ right collides with a stationary 1.0 kg block. After collision, the ball rebounds at 1 m s⁻¹ left. Find the block's velocity after collision.

1. 1. Take right as positive direction. List known values: $m_1 = 0.5$ kg, $u_1 = 4$ m s⁻¹, $v_1 = -1$ m s⁻¹, $m_2 = 1.0$ kg, $u_2 = 0$.
2. 2. Apply conservation of momentum:
3. $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$
4. 3. Substitute values:
5. $$(0.5 \times 4) + 0 = (0.5 \times -1) + 1.0 v_2$$
6. 4. Simplify and solve:
7. $$2 = -0.5 + v_2 \implies v_2 = 2.5 \text{ m s}^{-1} \text{ right}$$

## Common pitfalls

- **Wrong:** Adding a 'force of motion' to a free-body diagram
  - Why it fails: Objects keep moving due to inertia, not a continuous applied force
  - Correct: Only include actual interaction forces (weight, tension, friction etc.)
- **Wrong:** Mixing up the components of weight on an incline
  - Why it fails: Most students confuse which angle corresponds to which component
  - Correct: For incline angle $\theta$ to horizontal: parallel = $mg\sin\theta$, perpendicular = $mg\cos\theta$
- **Wrong:** Forgetting momentum is a vector and ignoring direction
  - Why it fails: Rebound velocity has opposite sign, which changes the final result
  - Correct: Always assign a positive direction before starting momentum calculations
- **Wrong:** Adding Newton's third law reaction force to the FBD
  - Why it fails: Reaction force acts on the other body, not the one you are analyzing
  - Correct: Only draw forces that act on your chosen object
- **Wrong:** Applying conservation of momentum to systems with external forces
  - Why it fails: Momentum is only conserved for isolated systems with zero net external force
  - Correct: Check for external forces like friction before applying the rule

## Cheatsheet

| Concept | Formula | Key Note |
| --- | --- | --- |
| Newton's Second Law | $F_{net} = ma$ | Vector, sum all components |
| Translational Equilibrium | $\sum F_x = 0, \sum F_y = 0$ | Acceleration = 0 |
| Momentum | $p = mv$ | Vector, units kg m s⁻¹ |
| Impulse | $J = F_{avg} \Delta t = \Delta p$ | Change in momentum |
| Conservation of Momentum (isolated) | $\sum p_{before} = \sum p_{after}$ | Use signs for direction |
| Weight on incline ($\theta$) | Parallel: $mg\sin\theta$, Perpendicular: $mg\cos\theta$ | Memorize this pair! |

## What's next

Forces and momentum form the foundation of all classical mechanics in IB Physics, and concepts from this subtopic appear in every other unit on the syllabus. Mastery of free-body diagrams and force resolution is critical for solving problems ranging from circular motion and gravitation to energy transfer and nuclear collisions. Conservation of momentum is a core physical rule that applies even to quantum and nuclear systems you will encounter later. Extend your knowledge next with work, energy and power, which connects force concepts to energy transfer, before moving on to circular motion and nuclear physics.

- [Work, energy and power](https://www.owlsprep.com/study/ib-physics-sl-u1-work-energy-and-power/)
- [Circular motion and gravitation](https://www.owlsprep.com/study/ib-physics-sl-u1-circular-motion-and-gravitation/)
- [The particulate nature of matter](https://www.owlsprep.com/study/ib-physics-sl-u2-overview/)

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