IB MYP Sciences – Forces Revision Notes

These revision notes condense forces for IB MYP Sciences into what you need in the final weeks. They are aligned to the International Baccalaureate Organization, Middle Years Programme Subject Brief – Sciences, from 2014, which lists forces among the topics explored in the MYP sciences on-screen examinations. They suit MYP years 4 and 5, and there is no SL/HL split in MYP sciences.

MYP has no prescribed content list: schools design their own units, and this page covers a topic the brief names. For full explanations and longer worked examples, go back to the forces study guide. When you are ready, try the forces practice questions. The course hub and printable checklist cover the rest of the course.

Where forces sits in the assessment

From the brief: the MYP sciences on-screen examination has three tasks.

Task	Criteria	Marks
Knowing and understanding	A	25
Investigation skills	B and C	50
Applying science	D	25

So forces can come up as recall and calculation (A), inside an experiment you design or data you process (B, C), or in a real-life issue you evaluate (D). Each criterion has eight achievement levels (1–8) in four bands. Your teacher will share the task-specific clarifications for your own school’s work. For more on the criteria, see the assessment revision notes.

Definitions

• Force – a push or pull; a vector measured in newtons (N).
• Contact forces – friction, air resistance (drag), tension, normal (reaction) force, upthrust.
• Non-contact forces – gravitational (weight), magnetic, electrostatic.
• Mass – amount of matter, in kg; the same everywhere.
• Weight – gravitational force on a mass, in N; depends on g.
• Resultant force – the single force equal in effect to all the forces acting.
• Speed – distance per unit time (scalar).
• Velocity – speed in a given direction (vector).
• Acceleration – rate of change of velocity.
• Terminal velocity – the constant velocity reached when drag equals weight.
• Pressure – force per unit area, acting at right angles to a surface.
• Moment – turning effect of a force about a pivot.

Equations

Quantity	Equation	Units
Weight	W = m × g	N (m in kg, g in N/kg)
Newton’s second law	F = m × a (F is the resultant)	N
Speed	speed = distance / time	m/s
Acceleration	a = (v − u) / t	m/s²
Pressure	p = F / A	Pa = N/m²
Pressure in a liquid	p = h × ρ × g	Pa
Moment	moment = F × d (d perpendicular to the force)	N m

Take g = 9.8 N/kg near Earth unless a question gives another value.

Newton’s laws in one line each

1. No resultant force → no change in motion (stays at rest or constant velocity).
2. Resultant force → acceleration in the direction of the force, with F = ma.
3. A on B means B on A: equal size, opposite direction, same type, different objects.

Method in steps

Resultant force and acceleration

1. Draw a free-body diagram of the one object.
2. Choose a positive direction.
3. Add the forces with signs to get the resultant.
4. Use a = F / m with the resultant only.
5. Give the direction with the answer.

Velocity-time graph questions

1. Gradient of each straight section = acceleration.
2. Split the area under the line into triangles and rectangles.
3. Add the areas for total distance.
4. Average speed = total distance ÷ total time (not the mean of two speeds).

Principle of moments

1. Mark the pivot.
2. For each force, find the perpendicular distance to the pivot.
3. Sum of clockwise moments = sum of anticlockwise moments.
4. Solve for the unknown and check units (m or N).

Small worked reminders

• A 1200 kg car with 3000 N driving force and 600 N resistance: resultant 2400 N, a = 2400 / 1200 = 2.0 m/s².
• A tram: 0 → 12 m/s in 4 s, 12 m/s for 6 s, 12 → 0 m/s in 3 s. Distance = 24 + 72 + 18 = 114 m.
• 600 N on 0.030 m²: p = 20 000 Pa. Same 600 N on 0.24 m²: p = 2500 Pa.
• Seesaw: 300 N at 2.0 m balances 400 N at d: 300 × 2.0 = 400d, so d = 1.5 m.

Graph shapes to recognise

Graph	Flat line	Straight sloping line	Curve getting steeper
Distance-time	Stationary	Constant speed	Speeding up
Velocity-time	Constant velocity	Constant acceleration	Increasing acceleration

On a velocity-time graph a line sloping down towards zero means deceleration. The area under it still counts as distance travelled.

Must-know distinctions

• Mass vs weight. Mass in kg never changes; weight in N changes with g. A 60 kg astronaut weighs 588 N on Earth and about 96 N on the Moon (g = 1.6 N/kg), but has a mass of 60 kg in both places.
• Speed vs velocity. Constant speed round a bend is a changing velocity, so there is an acceleration and a resultant force (towards the centre of the bend).
• Balanced forces vs no motion. Balanced forces mean constant velocity, which can be zero or not.
• Third-law pair vs balanced pair. A third-law pair acts on two different objects. Two balanced forces act on one object.
• Pressure vs force. A small force can make a large pressure if the area is tiny (a drawing pin).
• Moment vs force. Moment depends on distance from the pivot, so a small force far away can balance a large force close in.

Links to related concepts

The brief gives movement, energy and models as examples of MYP sciences related concepts. In a Criterion D answer you can use them to structure your thinking: seat belts and crumple zones (movement: they lengthen the time taken to stop), tyre design (models of pressure), or cycle helmets (energy absorbed by the foam).

Quick self-test

Use g = 9.8 N/kg. Give answers to 3 significant figures where they do not come out exactly.

1. Find the weight of a 2.5 kg bag on Earth.
2. Forces of 12 N to the right and 5 N to the left act on a box. Find the resultant.
3. A resultant force of 4.0 N acts on a 0.50 kg ball. Find its acceleration.
4. A cyclist travels 3.6 km in 12 minutes. Find the average speed in m/s.
5. A car slows from 25 m/s to 10 m/s in 5.0 s. Find its acceleration.
6. A 150 N force acts on an area of 0.030 m². Find the pressure.
7. A force of 18 N acts 0.40 m from a pivot, at right angles to the lever. Find the moment.
8. A 6.0 N weight hangs 0.30 m left of a pivot. How far right must a 4.0 N weight hang to balance it?
9. What does a horizontal line on a distance-time graph show?
10. A book rests on a table. What force forms a Newton’s third-law pair with the book’s weight?
11. A velocity-time graph rises in a straight line from 0 to 8 m/s in 4 s. How far does the object travel?
12. Find the extra pressure due to fresh water (ρ = 1000 kg/m³) at a depth of 2.0 m.

Answers

1. W = 2.5 × 9.8 = 24.5 N
2. 12 − 5 = 7 N to the right
3. a = 4.0 / 0.50 = 8.0 m/s²
4. 3600 m / 720 s = 5.0 m/s
5. a = (10 − 25) / 5.0 = −3.0 m/s² (a deceleration of 3.0 m/s²)
6. p = 150 / 0.030 = 5000 Pa
7. 18 × 0.40 = 7.2 N m
8. 6.0 × 0.30 = 4.0 × d, so d = 0.45 m
9. The object is stationary (speed is zero).
10. The gravitational pull of the book on the Earth (upwards, on the Earth). The table’s push is not the pair.
11. Area = ½ × 4 × 8 = 16 m
12. p = 2.0 × 1000 × 9.8 = 19 600 Pa

Where marks are usually lost

• Putting the driving force, not the resultant force, into F = ma.
• Leaving time in minutes or distance in km when the answer must be in m/s.
• Dropping the minus sign on a deceleration, or giving “−3.0 m/s² deceleration” (a double negative).
• Taking the height of a velocity-time graph as the distance instead of the area under it.
• Averaging the start and end speeds to get average speed over a journey with several phases.
• Leaving area in cm² when calculating pressure in pascals.
• Using the slanted length of a lever instead of the perpendicular distance in a moment.
• Naming the normal force as the third-law partner of weight.
• Describing terminal velocity as “no forces acting” rather than “forces balanced”.
• In an investigation, changing mass and force together so the effect of either cannot be separated.

Official syllabus

International Baccalaureate Organization, Middle Years Programme Subject Brief – Sciences, from 2014. Forces is one of the topics the brief lists for the MYP sciences on-screen examinations; task names, criteria and marks above are taken from the same brief.