Scalars, vectors and resolving vectorsEdexcel International A Level Physics: Revision notes

A scalar has magnitude only. A vector has magnitude and direction.

• Scalars: distance, speed, mass, time, energy, temperature
• Vectors: displacement, velocity, acceleration, force, weight, momentum

In print a vector is written in bold or with an arrow ($\vec{F}$); written by hand, with an underline or arrow. The magnitude of $\vec{F}$ is written $F$ or $|\vec{F}|$.

Distance is the total path length (scalar); displacement is the straight-line distance from start to end in a stated direction (vector). A hiker who walks 3.0 km east then 4.0 km north has travelled 7.0 km but has a displacement of 5.0 km.

A vector can be replaced by two components at right angles, usually horizontal and vertical (or parallel and perpendicular to a slope).

For a vector $F$ at angle $\theta$ to the horizontal:

• horizontal component $F_x = F\cos\theta$
• vertical component $F_y = F\sin\theta$

Example: a rope tension of 80 N at 30° above horizontal gives $80\cos30° = 69\ \text{N}$ horizontally and $80\sin30° = 40\ \text{N}$ vertically.

The component next to the angle uses cosine; the component opposite the angle uses sine. Components at right angles are independent: each has no effect along the other direction.

To resolve by drawing, choose a scale, draw the vector as a line to scale at the correct angle, then draw perpendicular lines from its ends to form a right-angled triangle. Measure the two sides and convert them back with the scale.

Example: for 80 N at 30°, with a scale of 1 cm to 10 N, draw an 8.0 cm line at 30°; the horizontal side measures 6.9 cm (69 N) and the vertical side 4.0 cm (40 N).

The resultant is the single vector with the same effect as two or more vectors. Add vectors head to tail so the resultant joins the start of the first to the end of the last.

For two vectors at right angles, calculate with Pythagoras and trigonometry: $R = \sqrt{A^2 + B^2}$ and $\tan\theta = \frac{B}{A}$.

Example: a boat heading at $3.0\ \text{m s}^{-1}$ across a river flowing at $1.5\ \text{m s}^{-1}$ has a resultant velocity of $\sqrt{3.0^2 + 1.5^2} = 3.4\ \text{m s}^{-1}$ at $\tan^{-1}(1.5/3.0) = 27°$ to the line straight across.

For vectors at any other angle, use a scale drawing: draw the first vector, draw the second from its end at the correct angle and measure the resultant's length and direction.