Vectors — resolving, adding, and the setups IB keeps reusing

Vectors are rarely examined on their own. They are examined inside every mechanics question, every fields question and a good deal of waves — quietly, as a skill the question assumes you already have. Getting a component right is usually the difference between a problem that takes two lines and one that will not come out at all...

Scalar or vector

SCALARS: distance, speed, mass, time, energy, force, work, power, temperature, charge ......

VECTORS: displacement, velocity, acceleration, momentum, magnetic field strength, impulse, gravitational ......

The pairs that catch people out are distance and displacement, speed and velocity, and mass and weight. 

If a question mentions an object returning to where it started, it is almost certainly testing that displacement is zero while distance is not.

Resolving into components

Any vector can be split into two perpendicular components. For a vector of magnitude F at an angle θ:

Resolving a force into components — the component adjacent to the angle takes cosine, the component opposite to the angle takes sine.

component adjacent to θ = F cos θ

component opposite to θ = F sin θ

Choose your axes to suit the problem rather than the page. On a slope, tilting the axes to run along and perpendicular to the surface turns an awkward problem into an easy one.

Do not memorise "horizontal is cos". That is only true when θ happens to be measured from the horizontal, and IB regularly measures it from somewhere else specifically to catch that. Look at where the angle actually sits: the component next to the angle takes cosine, the one across from it takes sine. This one habit removes most of the sign and function errors in the whole topic.

Adding vectors

For two vectors, draw them head to tail (or complete the parallelogram) and the resultant runs from the start of the first to the end of the second.

Head to tail or parallelogram — the resultant runs from the start of the first vector to the end of the last.

For any number of vectors, use components — it is faster and far less error-prone:

1. Resolve every vector into x and y components.
2. Add all the x components; add all the y components.
3. Recombine.

R = √ ( Rx² + Ry² )
θ = tan⁻¹ ( Ry / Rx )

Then check the quadrant. A calculator's tan⁻¹ only ever returns an angle between −90° and +90°, so if your resultant points into the second or third quadrant the calculator will quietly give you the wrong one. Sketching the resultant roughly before you compute it catches this every time.

Example 1: Three forces act at a point: 5.0 N at 0°, 4.0 N at 120° and 3.0 N at 230° (all measured from the +x axis). Find the resultant.

Three forces on one object. Resolve each onto the axes, add the components, then recombine.

Resolve every vector into x and y components:

F1x = 5.0 cos ( 0 ) = 5.0 N, F1y = 5.0 sin ( 0 ) = 0 N

F2X = 4.0 cos ( 120 ) = − 2.0 N, F2y = 4.0 sin ( 120 ) = 3.46 N

F3X = 3.0 cos ( 230 ) = − 1.928 N, F3y = 3.0 sin ( 230 ) = − 2.928 N

Add all the x components; add all the y components:

Fx = F1x + F2x + F3x = 5.0 − 2.0 − 1.928 = 1.072 N

Fy = F1y + F2y + F3y = 0 +3.46 − 2.928 =  1.166 N

Recombine:

F = √ ( Fx² + Fy²) = √ ( ( 1.072 )² + ( 1.166 )² ) = 1.58 N

θ = tan⁻¹ ( Fy / Fx ) = tan⁻¹ (  1.166 / 1.072 ) = 47.4° 

— both components positive, so the first quadrant ✓

Answer: 1.6 N at 47° from the +x axis.

The three setups that keep coming back

1 · Relative velocity.

Relative velocity is a subtraction — draw the triangle before you compute, and most of the errors disappear.

The velocity of A relative to B is a subtraction:

VAB = VA − VB

River crossings and aircraft in crosswinds are nothing more than this. Draw the triangle before you reach for the calculator; almost every mistake in these questions is a drawing mistake, not an arithmetic one.

2 · The inclined plane. 

Rotate your axes so that x runs down the slope. The weight Fg then resolves into Fg sin θ down the slope and Fg cos θ into the surface.

Rotate your axes so that x runs down the slope, and the weight resolves cleanly:

Along the slope: Fg sin θ. Perpendicular: Fg cos θ.

On a slope the normal force has magnitude Fg cos θ, not Fg — which is what rotating the axes makes obvious.

3 · Projectile motion. 

Resolve the launch velocity into u cos θ horizontally and u sin θ vertically, and the two components then behave completely independently — the horizontal one constant, the vertical one accelerating. Nothing passes between them except the time, which is why time is usually the variable to find first.

Where marks are actually lost

• Using sine where cosine belongs, because the angle was measured from an unfamiliar axis.
• Taking the normal force on a slope as Fg.
• Adding vector magnitudes as though they were scalars.
• Trusting tan⁻¹ without checking the quadrant.
• Setting acceleration to zero at the top of a projectile's path.
• Answering with a magnitude when the question asked for a vector, and so leaving out the direction.

Notes and practice

Next

Units, prefixes and scientific notation ➡️

Significant figure ➡️

Uncertainties and error analysis ➡️                                   

Paper 1B: data analysis and graphs ➡️

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