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Some physical quantities are fully specified by a single number and a unit: "the mass is 3.0 kg", "the temperature is 298 K". These are called scalars. Others require a direction as well as a magnitude to be meaningful: "a force of 20 N to the east", "a velocity of 5 m s⁻¹ downwards". These are called vectors. Distinguishing the two, and knowing how to combine vectors properly, is the foundation of all mechanics and fields work in the A-Level Physics course.
OCR's Module 2 specification requires you to: define scalar and vector; classify common quantities; add vectors using the tip-to-tail ("triangle") method; and combine two perpendicular vectors using Pythagoras and trigonometry. Lesson 10 will cover the inverse operation — resolving a vector into components — in detail.
A scalar quantity is one that has only magnitude (size) and, where appropriate, a unit. Adding two scalars is straightforward arithmetic: 3 kg + 2 kg = 5 kg. Scalars have no direction — it makes no sense to ask "which way does the mass point?".
| Quantity | Symbol | SI Unit |
|---|---|---|
| Mass | m | kg |
| Length, distance | l, d | m |
| Time | t | s |
| Temperature | T | K |
| Speed | v | m s⁻¹ |
| Energy, work | E, W | J |
| Power | P | W |
| Pressure | p | Pa |
| Electric charge | Q | C |
| Electric potential | V | V |
| Density | ρ | kg m⁻³ |
| Frequency | f | Hz |
Note that speed is a scalar, even though velocity is a vector. Speed is the magnitude of the velocity vector; it tells you how fast you are moving but not in which direction.
Similarly, distance is a scalar (the total path length travelled) while displacement is a vector (the straight-line change in position).
A vector quantity has both magnitude and direction. It is customarily drawn as an arrow: the length of the arrow represents the magnitude, and the direction of the arrow represents the physical direction of the quantity.
| Quantity | Symbol | SI Unit |
|---|---|---|
| Displacement | s, x | m |
| Velocity | v | m s⁻¹ |
| Acceleration | a | m s⁻² |
| Force | F | N |
| Momentum | p | kg m s⁻¹ |
| Electric field strength | E | N C⁻¹ |
| Magnetic flux density | B | T |
| Gravitational field strength | g | N kg⁻¹ |
| Weight | W | N |
Note that weight is a vector — it is a force, and forces always act in a particular direction (usually downward, for weight).
Vectors can be written in several equivalent ways:
Exam Tip: In handwritten answers, underline vectors (since bold is hard to write clearly). In typed answers, bold is preferred. OCR accepts either as long as it is clear you understand the distinction.
Imagine walking 3 km east, then 4 km north.
A car travels 60 km in 1 hour around a circular racetrack and ends up back at its starting point.
This is not a trick; it reflects a genuinely important distinction. A car that returns to its start has zero average velocity but a very non-zero average speed.
Two vectors are added by drawing the first, then drawing the second starting from where the first ended ("tip to tail"). The resultant vector is the arrow drawn from the very start of the first to the very tip of the second — this closes the triangle.
graph LR
A((Start)) -->|"Vector A"| B((Intermediate))
B -->|"Vector B"| C((End))
A -.->|"Resultant A + B"| C
The resultant has the magnitude of the closing side of the triangle and the direction from start to end.
A boat sails at 3 m s⁻¹ due east across a river flowing south at 4 m s⁻¹. Find its resultant velocity (magnitude and direction relative to east).
Solution:
Draw a right-angled triangle with the 3 m s⁻¹ east as the horizontal side and the 4 m s⁻¹ south as the vertical side. The resultant is the hypotenuse.
Magnitude:
v = √(3² + 4²) = √(9 + 16) = √25 = 5 m s⁻¹
Direction (measured from east, towards south):
tan θ = 4 / 3 ⇒ θ = arctan(4/3) = 53.1°
So the resultant velocity is 5 m s⁻¹ at 53° south of east.
A box on a frictionless surface is pulled simultaneously by two ropes: 30 N north and 40 N east. Calculate the magnitude and direction of the resultant force.
Solution:
F = √(30² + 40²) = √(900 + 1600) = √2500 = 50 N θ = arctan(30/40) = arctan(0.75) = 36.9°
Resultant: 50 N at 37° north of east.
When the vectors are not perpendicular, you cannot use Pythagoras directly. You can either:
For A-Level, options 1 and 2 are most common. The cosine rule method is mathematically elegant but rarely required.
For two vectors A and B separated by an angle θ between their directions when drawn tip-to-tail (i.e. the external angle at their meeting point), the magnitude of the resultant is:
|R|² = |A|² + |B|² − 2|A||B| cos(180° − θ) = |A|² + |B|² + 2|A||B| cos θ
And the direction can be found using the sine rule.
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