# A.2 Forces and momentum

> IB Physics HL · IB Physics HL (First Assessment 2025)
> Source: https://www.owlsprep.com/study/ib-physics-hl-u1-a-2-forces-and-momentum/

This core mechanics sub-topic covers Newton's laws of motion, linear momentum, impulse, and conservation of momentum. You will learn to solve force, collision, and explosion problems that appear frequently in IB Physics exams.

**Prerequisites:** [A.1 Kinematics](https://www.owlsprep.com/study/ib-physics-hl-u1-a-1-kinematics/); Vector basics

## Learning objectives

- Apply Newton's three laws of motion to solve force problems in 1 dimension
- Calculate impulse and change in momentum for linear systems
- Use conservation of momentum to solve collision and explosion problems
- Distinguish between elastic and inelastic collisions based on kinetic energy conservation

## Newton's Three Laws of Motion

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

**Newton's Second Law** — The rate of change of momentum of a body equals the net external force acting on it:

*Example:* For constant mass, this simplifies to $\vec{F}_{net} = m\vec{a}$.

**Newton's Third Law** — If body A exerts a force on body B, body B exerts an equal magnitude, opposite direction force of the same type on body A.

**Worked example:** A 5.0 kg box accelerates at $2.0 \text{ m s}^{-2}$ along a horizontal surface against a constant 3.0 N frictional force. Calculate the applied force on the box.

1. 1. Draw a free-body diagram and apply Newton's second law along the horizontal axis:
2. $$F_{applied} - F_{friction} = ma$$
3. 2. Substitute the given values to solve for $F_{applied}$:
4. $$F_{applied} = ma + F_{friction} = (5.0 \times 2.0) + 3.0 = 13 \text{ N}$$

> **Exam tip:** Always draw a free-body diagram and label all forces before applying Newton's second law to avoid missing forces.

## Momentum and Impulse

**Linear Momentum** — A vector quantity equal to the product of an object's mass and velocity: $\vec{p} = m\vec{v}$.

*Notation:* p

**Impulse** — Impulse equals the change in momentum of an object, given by $J = F_{avg} \Delta t = \Delta p$. Impulse is also equal to the area under a force-time graph.

*Notation:* J

**Worked example:** A 0.15 kg ball hits a wall horizontally at $10 \text{ m s}^{-1}$, and rebounds straight back at $8 \text{ m s}^{-1}$. Calculate the impulse exerted on the ball by the wall.

1. 1. Assign the initial direction of the ball as positive, so $v_i = +10 \text{ m s}^{-1}$ and $v_f = -8 \text{ m s}^{-1}$:
2. 2. Calculate change in momentum, which equals impulse:
3. $$\Delta p = m(v_f - v_i) = 0.15(-8 - 10) = -2.7 \text{ kg m s}^{-1}$$
4. The negative sign indicates impulse acts opposite to the ball's initial direction, so impulse on the ball is $-2.7 \text{ N s}$.

## Conservation of Momentum

**Conservation of Linear Momentum** — For a closed system (no mass enters or leaves) with zero net external force, the total momentum of the system before an interaction equals the total momentum after the interaction: $\sum \vec{p}_{initial} = \sum \vec{p}_{final}$.

**Worked example:** A 70 kg person stands at rest on a stationary 100 kg free-floating boat. If the person walks at $1.5 \text{ m s}^{-1}$ relative to water, what is the boat's velocity relative to water?

1. 1. Initial total momentum of the system (person + boat) is 0, since both are stationary:
2. 2. Apply conservation of momentum, let $v_b$ = boat velocity:
3. $$m_p v_p + m_b v_b = 0$$
4. 3. Solve for $v_b$:
5. $$v_b = -\frac{m_p v_p}{m_b} = -\frac{70 \times 1.5}{100} = -1.05 \text{ m s}^{-1}$$
6. The negative sign means the boat moves opposite to the person's direction.

> **Exam tip:** Always confirm there is no net external force on the system before applying conservation of momentum.

## Types of Collisions

- **Elastic collisions**: Both total momentum *and* total kinetic energy are conserved
- **Inelastic collisions**: Only total momentum is conserved; kinetic energy is lost to heat, sound, or deformation
- **Perfectly inelastic collisions**: Objects stick together after collision, with maximum kinetic energy loss

**Worked example:** A 2.0 kg mass moving at $3.0 \text{ m s}^{-1}$ collides head-on with a stationary 1.0 kg mass. After collision, the 2.0 kg mass moves at $1.0 \text{ m s}^{-1}$ in the same direction. Is the collision elastic?

1. 1. Use conservation of momentum to find the 1.0 kg mass's final velocity:
2. $$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$$
3. $$(2.0 \times 3.0) + 0 = (2.0 \times 1.0) + 1.0 v_{2f} \implies v_{2f} = 4.0 \text{ m s}^{-1}$$
4. 2. Compare total kinetic energy before and after collision:
5. $$KE_{initial} = \frac{1}{2} \times 2.0 \times 3^2 = 9.0 \text{ J}$$
6. $$KE_{final} = (\frac{1}{2} \times 2.0 \times 1^2) + (\frac{1}{2} \times 1.0 \times 4^2) = 1.0 + 8.0 = 9.0 \text{ J}$$
7. Kinetic energy is conserved, so the collision is elastic.

## Common pitfalls

- **Wrong:** Forgetting momentum is a vector and ignoring direction signs
  - Why it fails: Adding magnitudes of momentum for opposite directions gives incorrect change in momentum
  - Correct: Always assign a positive direction before calculation, and keep velocity and momentum signs consistent
- **Wrong:** Confusing Newton's third law force pairs with balanced forces on one body
  - Why it fails: Force pairs act on *different* bodies, so they do not cancel out for a single body
  - Correct: Draw separate free-body diagrams for each body, and label which body each force acts on
- **Wrong:** Assuming kinetic energy is conserved in all collisions
  - Why it fails: Only elastic collisions conserve kinetic energy, which is rare for macroscopic collisions
  - Correct: Only use kinetic energy conservation if the question explicitly states the collision is elastic, or you confirm it via calculation
- **Wrong:** Applying conservation of momentum when there is a net external force
  - Why it fails: The impulse from the external force changes the system's total momentum
  - Correct: Expand the system to include the object exerting the external force, or account for the impulse from the external force

## Cheatsheet

| Concept | Formula | Key Notes |
| --- | --- | --- |
| Newton's 2nd Law (general) | $F_{net} = \frac{dp}{dt}$ | True for all mass cases |
| Momentum | $p = mv$ | Vector quantity |
| Impulse-Momentum | $J = F_{avg}\Delta t = \Delta p$ | Impulse = area under F-t graph |
| Conservation of Momentum | $\sum p_i = \sum p_f$ | Valid for closed systems, $F_{net,ext}=0$ |
| Elastic Collision | p and KE conserved | No net energy loss |
| Inelastic Collision | Only p conserved | Kinetic energy lost to other forms |

## What's next

Forces and momentum form the foundation of all classical mechanics in IB Physics HL, and questions on this topic appear in almost every exam paper. The concepts you learned here will be extended to circular motion, where you will apply Newton's second law to centripetal force problems, and to work and energy, where you will combine momentum and energy conservation to solve complex multi-step problems. You will also use momentum to analyze particle interactions in nuclear physics later in the course, so mastering this sub-topic is critical for success in higher-level topics. Below are related sub-topics to study next:

- [A.3 Work, energy and power](https://www.owlsprep.com/study/ib-physics-hl-u1-a-3-work-energy-and/)
- [A.4 Rotational mechanics](https://www.owlsprep.com/study/ib-physics-hl-u1-a-4-rotational-mechanics/)
- [A.5 Special relativity: energy and momentum (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u1-a-5-special-relativity-energy/)

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