AQA AS Level Mathematics Paper 1, June 2025: Question 19

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Q Marking instructions AO Marks Typical solution

19 2 ( )2 1.1b B1 Magnitude of given vector

Obtains p + −0 5. or

2 ( )2

22 = p + −0 5.

p + 0 5.

ACF

Condone any variable for p p2 + 0 25. = 1.3

Equates their expression for the 3.1a M1 2

p = 1.44

magnitude of the given vector to

1.3 or the square of the

p = ±1.2

magnitude of the given vector to

1.32 to obtain at least one value

for p

Deduces p = ±1.2 2.2a R1

Question 19 Total 3

How to answer it

Finding Unknown Vector Components from Magnitude

📋 What this question tests

This question evaluates your core mechanics and vector algebra skills, specifically:

Applying Pythagoras' Theorem to calculate the magnitude of a 2D column vector.
Setting up and solving a quadratic equation involving an unknown component.
Recognising that squaring creates two symmetrical solutions: both positive and negative roots.

Question 19

Vectors in Mechanics • 3 Marks Total

💡 Key Knowledge

For any 2D vector F = [x, y], the magnitude is defined as:
|F| = √(x² + y²)
Squaring both sides gives:
|F|² = x² + y²
Squaring a negative quantity always gives a positive value:
(-0.5)² = +0.25
When solving an equation of the form p² = k (where k > 0), there are two real solutions: p = ±√k .

🧠 Exam Technique & Clues

Notice plural keywords: The question asks for the "possible values" of p. This is a direct exam hint that there is more than one answer!
Work with squared magnitudes: Avoid messy square roots by equating p² + (-0.5)² = 1.3² directly.
Look for Pythagorean triples: 5, 12, 13 scaled down by 10 gives 0.5, 1.2, 1.3. This allows quick mental verification.

📐 Step-by-Step Solution

Write down the expression for magnitude:
|F| = √(p² + (-0.5)²) = √(p² + 0.25)
[B1] Awarded for writing p² + (-0.5)² or p² + 0.5² (any equivalent form).
Equate to given magnitude and solve for p:
√(p² + 0.25) = 1.3
Square both sides:
p² + 0.25 = 1.3²
p² + 0.25 = 1.69
p² = 1.69 - 0.25 = 1.44
[M1] Awarded for equating the expression (or squared expression) to 1.3 (or 1.3²) to obtain at least one value for p.
Deduce all valid solutions:
p = ±√1.44 = ±1.2 (or p = ±6/5 )
[R1] Reasoning mark awarded strictly for deducing both p = ±1.2 .

✅ Final Correct Answer

The possible values of p are:

p = 1.2 or p = -1.2

Equivalent fraction forms like ±6/5 or ±1 1/5 are fully accepted.

❌ Common Errors & Lost Marks

Losing the negative root: Writing only p = 1.2 and omitting p = -1.2 is the single most common reason students drop the final R1 mark.
Sign errors during squaring: Calculating (-0.5)² as -0.25 , leading to p² - 0.25 = 1.69 ⇒ p² = 1.94 .
Premature rounding: Always maintain exact decimals here since 1.3² = 1.69 and 1.44 are terminating numbers.

💬 Examiner Insight

This question tests standard procedural fluency (AO1.1b), mathematical problem solving (AO3.1a), and mathematical deduction (AO2.2a). While the majority of students comfortably set up the Pythagoras equation and scored the first two marks, many missed the final deduction mark R1 by stopping at p = 1.2 . In mechanics vectors, components represent signed directional quantities—never discard negative values unless the question explicitly states a geometric constraint such as p > 0 .

Topics

Mechanics · Pure Mathematics · J: Vectors · R: Forces and Newton’s laws