Scalars, vectors and resolvingEdexcel A-Level Physics: Revision notes

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

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

In print a vector is written in bold (v) or with an arrow or underline; its magnitude is |v| or just v. Add direction by a bearing, an angle to a stated line, or a sign along a line. Distance is the total path length; displacement is the straight-line change in position, so for any journey displacement ≤ distance.

The resultant is the single vector with the same effect as two or more vectors acting together. Vectors are added by placing them tip to tail: the resultant joins the tail of the first to the tip of the last. This is the triangle rule. Equivalent is the parallelogram rule: draw both vectors from one point, complete the parallelogram, and the diagonal from that point is the resultant.

Order does not matter (a + b = b + a). To subtract a vector, add its reverse: a − b = a + (−b).

This method works for any angle. Choose a convenient scale (for example 1 cm to 5 N), and draw the first vector from a point using a protractor for its direction. Draw the second from the tip of the first. Join the start to the finish: measure the length with a ruler and the angle with a protractor, then convert the length using the scale.

Accuracy is limited by drawing and reading errors, so use a sharp pencil, the largest scale that fits, and state the answer to a sensible precision (usually 2 or 3 significant figures).

When two vectors are at right angles the triangle is right-angled, so use Pythagoras and trigonometry:

• Magnitude: R = √(a² + b²)
• Angle to vector a: tan θ = b/a

Worked example. A boat heads at 3.0 m s⁻¹ across a river that flows at 1.5 m s⁻¹. R = √(3.0² + 1.5²) = 3.35 m s⁻¹, and the angle to the heading is tan⁻¹(1.5/3.0) = 26.6° downstream.

A vector F at angle θ to a reference direction can be replaced by two perpendicular components: F cos θ along the reference line and F sin θ at right angles to it. Taken together they have the same effect as the original.

For a force at 35° above the horizontal: horizontal component = F cos 35°, vertical component = F sin 35°. If the angle is measured from the vertical, the sine and cosine swap. Check the sense of your answer: a larger angle from the reference direction reduces the component along it.

Resolving is also used for slopes, where a weight W splits into W sin θ down the slope and W cos θ perpendicular to it.

For vectors that are not at right angles, resolve each into components along two perpendicular directions, add the components along each direction, then combine the totals with Pythagoras and trigonometry.

Worked example. 300 N due north and 400 N at 60° to it on the east side. North: 300 + 400 cos 60° = 500 N. East: 400 sin 60° = 346 N. R = √(500² + 346²) = 608 N, at tan⁻¹(346/500) = 35° east of north.

This is more precise than a scale diagram, which can be used to check it.