AQA A-Level Mathematics Paper 2, June 2025: Question 15

8 marks · Medium difficulty · Multi-step Problem

Find the resultant of two forces given by magnitudes and bearings, calculate the angle it makes with one force, and determine the magnitude and bearing of an equilibrant force.

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Question

Question text

15 A particle moves under the actions of two forces, F1 and F2

F1 has magnitude 17 newtons and acts due East.

F2 has magnitude 26 newtons and acts at a bearing of 310°

The resultant of F1 and F2 is R

15 (a) Show that the magnitude of R is 17.0 newtons, correct to three significant figures.

[4 marks]

15 (b) Find the angle that R makes with F1

Give your answer to the nearest degree.

[2 marks]

… 23

(22)

15 (c) A third force, F3, acts upon the particle so that the particle is in equilibrium.

15 (c) (i) State the magnitude of F3

[1 mark]

15 (c) (ii) State the bearing on which F3 acts.

[1 mark]

Mark scheme

Show the mark scheme

Q Marking instructions AO Marks Typical solution

15(a) Resolves horizontally or 3.3 M1

vertically to find a correct

expression for one of the i or j Horizontally: 17 − 26cos(40)

components of the resultant

= −2 917 …

force.

Or Vertically: 26sin(40)

Uses the cosine rule with b = 17 = 16 712 …

and c = 26

(−2 917.)2 +16 712.2 = 16 97.N

Resolves horizontally and 1.1b A1

vertically to find a correct

17 0.N

expression for both i and j

components of the resultant

force

PI by H: AWRT 2 9.

V: AWRT 16 7.

Uses the cosine rule to obtain

a2 = AWRT 287.8

Finds the magnitude of the 1.1a M1

resultant vector of their

horizontal and vertical

components

Finds the square root of their

expression for b2 + c2 –2bcCosA

PI by AWFW [16.96, 16.97]

Completes reasoned argument

showing a full method to obtain

AWFW [16.96, 16.97] and

concludes that the magnitude of 2.1 R1

R is 17.0

Condone 17 – A-LEVEL MATHEMATICS – 7357/2 –

Subtotal 4

15(b) Uses an appropriate 3.1a M1

trigonometric equation using −1 16 712. o

their components for R tan = 80

2 917.

180 − 80 = 100o

Deduces that the two 17N

forces make this an isosceles

triangle, eg: 180 – (2 x 40)

PI AWRT 79 or 80

Obtains any of the following 1.1b A1

angles

AWRT 100, 101, 259, 260

Subtotal 2

15(c)(i) States 17.0 2.2a B1 17.0 N

PI by AWRT 17

Subtotal 1

15(c)(ii) States 170° 170°

2.2a B1

PI by AWRT 170

Subtotal 1

Question 15 Total 8

How to answer it

Resultant Forces, Bearings & Equilibrium

📌 What this question tests

This question assesses your ability to model coplanar forces using vectors and trigonometry. Specifically, you need to:

Resolve vectors into orthogonal horizontal (East-West) and vertical (North-South) components from bearings.
Combine perpendicular components using Pythagoras' Theorem to show the magnitude of a resultant force.
Alternative method: Apply the Cosine Rule directly to a closed vector triangle.
Calculate directional angles between vectors using trigonometry.
Apply the conditions of static equilibrium (resultant force = 0) to find the magnitude and bearing of an equilibrant force.

Part (a) — 4 Marks

Show Magnitude of Resultant R is 17.0 N

Required: Rigorous calculation showing the unrounded value before concluding 17.0 N

📐 Method 1: Resolving into Components (i and j)

Step 1: Understand bearings & directions
• F₁ = 17 N due East → 17 i + 0 j
• F₂ = 26 N on bearing 310° (i.e. 50° West of North, or 40° North of West).
Horizontal component of F₂ = -26 sin(50°) = -26 cos(40°) = -19.918 N
Vertical component of F₂ = +26 cos(50°) = +26 sin(40°) = +16.712 N

Step 2: Sum components to find R
Horizontal (i): 17 - 26 cos(40°) = -2.917... N
Vertical (j): 26 sin(40°) = +16.712... N

Step 3: Magnitude via Pythagoras
|R| = √((-2.917...)² + (16.712...)²)
|R| = √(8.510 + 279.30) = √287.81... = 16.965... N

Step 4: Conclude clearly
16.965... N = 17.0 N (to 3 s.f.) [as required]

📐 Method 2: Cosine Rule in Vector Triangle

Step 1: Angle between vectors
Due East is bearing 090°. The direction of F₂ is 310°.
Angle between their tails = 360° - 310° + 90° = 140°.
In a tip-to-tail triangle, the interior angle opposite R is:
180° - 140° = 40°.

Step 2: Apply the Cosine Rule
|R|² = 17² + 26² - 2(17)(26)cos(40°)
|R|² = 289 + 676 - 884 cos(40°)
|R|² = 965 - 677.185... = 287.815...

Step 3: Square root & conclude
|R| = √287.815... = 16.965... N
= 17.0 N (to 3 s.f.)

🧠 Exam Technique: "Show that" Questions

Because the answer 17.0 is given in the question, the final mark is an R mark (reasoned argument).

You must write down an unrounded intermediate value (e.g. 16.96 or 16.97) before rounding to 17.0.
Jumping straight from the square root to 17.0 will lose the final mark.

❌ Common Errors

Sign error on horizontal component: Forgetting that bearing 310° points West (negative i), leading to adding 19.9 to 17 instead of subtracting.
Premature rounding: Rounding -2.917 and 16.712 to 1 s.f. too early will corrupt the final 3 s.f. accuracy.

Mark Scheme Breakdown:
• [M1] Resolves horizontally or vertically for one component OR sets up cosine rule with 17 and 26.
• [A1] Both components correct (H: awrt ±2.9, V: awrt ±16.7) OR cosine rule gives a² = awrt 287.8.
• [M1] Uses Pythagoras on their components OR takes square root of cosine rule expression.
• [R1] Complete, fully reasoned method reaching unrounded awfw [16.96, 16.97] and concluding 17.0 N.

Part (b) — 2 Marks

Find the Angle that R Makes with F₁

Required: Correct direction relative to East, rounded to the nearest degree

✅ Correct Answer

100°

Also accepted: 101°, ±259°, ±260°

💡 Geometric Insight (Isosceles Triangle)

Notice that |F₁| = 17 N and |R| ≈ 17.0 N! The vector triangle formed by F₁, F₂, and R has two equal sides of length 17 N.

Opposite angles are equal to 40°, meaning the angle at the peak is 180° - 2(40°) = 100°.

📐 Component Method Step-by-Step

Step 1: Find angle with the West axis
The horizontal component is -2.917 N (acting West) and vertical is +16.712 N (acting North).
θ = tan⁻¹(16.712 / 2.917) = 80.1° North of West.

Step 2: Adjust reference to F₁ (Due East)
Since F₁ points due East, the angle between East and R is:
Angle = 180° - 80.1° = 99.9° ≈ 100° (to nearest degree).

Mark Scheme Breakdown:
• [M1] Uses appropriate trigonometric equation (e.g. tan⁻¹(16.712/2.917) = 80°) OR recognises isosceles triangle 180 - (2 × 40).
• [A1] Correct angle to nearest degree: 100° (or 101°, ±259°, ±260°).

Part (c) — 2 Marks Total

Equilibrium with a Third Force F₃

When a system is in equilibrium: F₁ + F₂ + F₃ = 0 ⇒ F₃ = -R

(c)(i) State the magnitude of F₃ [1 Mark]

✅ Correct Answer

17.0 N (or 17 N)

💡 Why?

For the particle to remain in equilibrium, F₃ must have the exact same magnitude as the resultant R, but act in the directly opposite direction: |F₃| = |R| = 17.0 N.

• [B1] States 17.0 (or 17).

(c)(ii) State the bearing on which F₃ acts [1 Mark]

✅ Correct Answer

170°

📐 Bearing Calculation

1. F₁ is on bearing 090°.
2. R is rotated 100° anticlockwise from East: Bearing of R = 090° - 100° = -010° ≡ 350°.
3. F₃ acts directly opposite to R (opposite direction = add or subtract 180°):
Bearing of F₃ = 350° - 180° = 170°.

❌ Common Error: Giving the bearing of R instead of F₃

Students often correctly calculate 350° for the resultant R, but forget that an equilibrating force acts in the opposite direction ( F₃ = -R ), requiring a 180° shift.

• [B1] States 170° (accept awrt 170).

Topics

Mechanics · R: Forces and Newton’s laws

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.