Physics 9702 · Paper 2

Adding Vectors

Adding two vectors gives one combined vector, the .
Scalars add as plain numbers. Vectors do not: their directions matter, and the answer needs a magnitude and a direction.

At right angles: Pythagoras and tan

Draw the vectors head to tail: start the second vector at the tip of the first. The resultant runs from the very start to the very end.
When the two vectors are at right angles, the triangle is right-angled, so Pythagoras and tan give the answer directly.

R = \sqrt{a^2 + b^2} \qquad \theta = \tan^{-1}\!\left(\dfrac{b}{a}\right)

Symbols

$a,\ b$ = the two perpendicular vectors (their own unit)
$R$ = magnitude of the resultant (same as a and b)
$\theta$ = angle of the resultant, measured from a's direction (degrees)

Head to tail, drawn to scale: 3 N east then 4 N north. The side that closes the triangle is the resultant, R = 5 N at 53° north of east.

Worked example

Smallest case: two forces at right angles

A 3 N force points east and a 4 N force points north. Find the resultant.

Magnitude: $R = \sqrt{3^2 + 4^2} = 5 \text{ N}$ .
Direction, measured from east: $\theta = \tan^{-1}\!\left(\tfrac{4}{3}\right) = 53^\circ$ north of east.

Answer

R = 5 \text{ N at } 53^\circ \text{ N of E}

A common variant asks for the direction of $X - Y$ as well as $X + Y$ . Subtracting a vector means adding it reversed: turn $Y$ around by 180°, then add head to tail as normal. (9702/11/M/J/25 Q4)

Not at right angles: resolve, add, recombine

Pythagoras only works at 90°. For any other angle, use the components from the last page:

Resolve each vector into two perpendicular components (for example north and east).
Add the components in each direction separately, with signs.
Recombine: Pythagoras for the magnitude, tan for the direction.

Head to tail with real angles: the 13 m s⁻¹ ship velocity, then the 2.7 m s⁻¹ current due west. The side that closes the triangle is the resultant, about 12 m s⁻¹ at 24° east of north.

Worked example

Ship plus current

A ship moves at 13 m s⁻¹, 35° east of north. The water has a current of 2.7 m s⁻¹ due west. Find the ship's resultant velocity.

Resolve the ship's velocity: $v_N = 13\cos 35^\circ = 10.6$ , $v_E = 13\sin 35^\circ = 7.5 \text{ m s}^{-1}$ .
The current is due west, so it only changes the east component: $v_E = 7.5 - 2.7 = 4.8 \text{ m s}^{-1}$ .
Magnitude: $R = \sqrt{10.6^2 + 4.8^2} = 12 \text{ m s}^{-1}$ .
Direction from north: $\tan^{-1}\!\left(\tfrac{4.8}{10.6}\right) = 24^\circ$ east of north.

Answer

R = 12 \text{ m s}^{-1} \text{ at } 24^\circ \text{ E of N}

(9702/23/M/J/21 Q1(b))

The scale-drawing method

A vector triangle drawn to scale (say 1 cm to 1 N) is an accepted method: place the vectors head to tail and measure the closing side with a ruler and protractor.
Draw it large. On a large diagram, a small drawing error changes the answer less.
Resolving is usually faster and more precise, so use the scale drawing only when the question asks for it.

A vector answer is complete only with both numbers: a magnitude and a direction. Stopping at the magnitude loses a mark.

Common mistake

Two common errors: giving a magnitude with no direction, and using Pythagoras on vectors that are not at right angles. Check the angle between the vectors before choosing the method.

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Adding Vectors

Adding two vectors gives one combined vector, the .
Scalars add as plain numbers. Vectors do not: their directions matter, and the answer needs a magnitude and a direction.

At right angles: Pythagoras and tan

Draw the vectors head to tail: start the second vector at the tip of the first. The resultant runs from the very start to the very end.
When the two vectors are at right angles, the triangle is right-angled, so Pythagoras and tan give the answer directly.

R = \sqrt{a^2 + b^2} \qquad \theta = \tan^{-1}\!\left(\dfrac{b}{a}\right)

Symbols

$a,\ b$ = the two perpendicular vectors (their own unit)
$R$ = magnitude of the resultant (same as a and b)
$\theta$ = angle of the resultant, measured from a's direction (degrees)

Head to tail, drawn to scale: 3 N east then 4 N north. The side that closes the triangle is the resultant, R = 5 N at 53° north of east.

Worked example

Smallest case: two forces at right angles

A 3 N force points east and a 4 N force points north. Find the resultant.

Magnitude: $R = \sqrt{3^2 + 4^2} = 5 \text{ N}$ .
Direction, measured from east: $\theta = \tan^{-1}\!\left(\tfrac{4}{3}\right) = 53^\circ$ north of east.

Answer

R = 5 \text{ N at } 53^\circ \text{ N of E}

A common variant asks for the direction of $X - Y$ as well as $X + Y$ . Subtracting a vector means adding it reversed: turn $Y$ around by 180°, then add head to tail as normal. (9702/11/M/J/25 Q4)

Not at right angles: resolve, add, recombine

Pythagoras only works at 90°. For any other angle, use the components from the last page:

Resolve each vector into two perpendicular components (for example north and east).
Add the components in each direction separately, with signs.
Recombine: Pythagoras for the magnitude, tan for the direction.

Head to tail with real angles: the 13 m s⁻¹ ship velocity, then the 2.7 m s⁻¹ current due west. The side that closes the triangle is the resultant, about 12 m s⁻¹ at 24° east of north.

Worked example

Ship plus current

A ship moves at 13 m s⁻¹, 35° east of north. The water has a current of 2.7 m s⁻¹ due west. Find the ship's resultant velocity.

Resolve the ship's velocity: $v_N = 13\cos 35^\circ = 10.6$ , $v_E = 13\sin 35^\circ = 7.5 \text{ m s}^{-1}$ .
The current is due west, so it only changes the east component: $v_E = 7.5 - 2.7 = 4.8 \text{ m s}^{-1}$ .
Magnitude: $R = \sqrt{10.6^2 + 4.8^2} = 12 \text{ m s}^{-1}$ .
Direction from north: $\tan^{-1}\!\left(\tfrac{4.8}{10.6}\right) = 24^\circ$ east of north.

Answer

R = 12 \text{ m s}^{-1} \text{ at } 24^\circ \text{ E of N}

(9702/23/M/J/21 Q1(b))

The scale-drawing method

A vector triangle drawn to scale (say 1 cm to 1 N) is an accepted method: place the vectors head to tail and measure the closing side with a ruler and protractor.
Draw it large. On a large diagram, a small drawing error changes the answer less.
Resolving is usually faster and more precise, so use the scale drawing only when the question asks for it.

A vector answer is complete only with both numbers: a magnitude and a direction. Stopping at the magnitude loses a mark.

Common mistake

Two common errors: giving a magnitude with no direction, and using Pythagoras on vectors that are not at right angles. Check the angle between the vectors before choosing the method.