A.2 Forces and momentum

IB Physics HL· Theme A: Space, time and motion, Topic A.2· 25 min read

1. Newton's Three Laws of Motion★★☆☆☆⏱ 8 min

📘 Definition

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.

📘 Definition

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 $F_{n e t} = m a$ .

📘 Definition

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 $2.0 m s^{- 2}$ 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
1. Substitute the given values to solve for $F_{a ppl i e d}$ :
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.

2. Momentum and Impulse★★★☆☆⏱ 7 min

📘 Definition

Linear Momentum

p

A vector quantity equal to the product of an object's mass and velocity: $p = m v$ .

📘 Definition

Impulse

J

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

📐 Worked Example

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

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

3. Conservation of Momentum★★★☆☆⏱ 8 min

📘 Definition

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: $\sum p_{ini t ia l} = \sum p_{f ina l}$ .

📐 Worked Example

A 70 kg person stands at rest on a stationary 100 kg free-floating boat. If the person walks at $1.5 m s^{- 1}$ 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:
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1. Apply conservation of momentum, let $v_{b}$ = boat velocity:
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$m_p v_p + m_b v_b = 0$
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1. Solve for $v_{b}$ :
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$v_b = -\frac{m_p v_p}{m_b} = -\frac{70 \times 1.5}{100} = -1.05 \text{ m s}^{-1}$
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.

4. Types of Collisions★★★☆☆⏱ 6 min

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

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

5. Common Pitfalls

Wrong move:

Forgetting momentum is a vector and ignoring direction signs

Why:

Adding magnitudes of momentum for opposite directions gives incorrect change in momentum

Correct move:

Always assign a positive direction before calculation, and keep velocity and momentum signs consistent

Wrong move:

Confusing Newton's third law force pairs with balanced forces on one body

Why:

Force pairs act on different bodies, so they do not cancel out for a single body

Correct move:

Draw separate free-body diagrams for each body, and label which body each force acts on

Wrong move:

Assuming kinetic energy is conserved in all collisions

Why:

Only elastic collisions conserve kinetic energy, which is rare for macroscopic collisions

Correct move:

Only use kinetic energy conservation if the question explicitly states the collision is elastic, or you confirm it via calculation

Wrong move:

Applying conservation of momentum when there is a net external force

Why:

The impulse from the external force changes the system's total momentum

Correct move:

Expand the system to include the object exerting the external force, or account for the impulse from the external force

6. Quick Reference Cheatsheet

Concept	Formula	Key Notes
Newton's 2nd Law (general)	$F_{n e t} = \frac{d p}{d t}$	True for all mass cases
Momentum	$p = m v$	Vector quantity
Impulse-Momentum	$J = F_{a v g} Δ t = Δ p$	Impulse = area under F-t graph
Conservation of Momentum	$\sum p_{i} = \sum p_{f}$	Valid for closed systems, $F_{n e t, e x t} = 0$
Elastic Collision	p and KE conserved	No net energy loss
Inelastic Collision	Only p conserved	Kinetic energy lost to other forms

7. Frequently Asked

Is momentum conserved in all collisions?

Momentum is always conserved for closed systems with no net external force, regardless of collision type. Only kinetic energy conservation differs between elastic and inelastic collisions.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · Paper 1
Impulse from force-time graph
2024 · Paper 2
Conservation of momentum explosion
2023 · Paper 1
Newton's third law force pairs

Going deeper

syllabusIB Physics HL Official SyllabusTopic A.2 full syllabus specification

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: