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HSC-EXT2 · Mechanics

Mechanics

HSC HSC Mathematics Extension 2 · 2024 content registry · first HSC 2027

Official NSW Curriculum syllabus and implementation timeline ↗Download questions and solutions PDF
MEX-M1
Assessment evidence

What these examples cover

These examples show assessable skills, not an official HSC topic weighting or prediction. The 2024 NSW syllabuses have their first HSC examination in 2027; students sitting the 2026 HSC continue under the applicable 2017 syllabus.

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3 deterministic examples

Mechanics questions with worked solutions

Each question is generated from an existing curriq QDSL archetype at a fixed seed. Marks and catalog content mappings travel with the generated question.

Worked example 1

2 marksext2-mechanics-1
Acceleration from velocity as a function of position A particle has v=(3x+2)/(2x−1)v=(3x+2)/(2x-1) for x>=1x>=1. Find its acceleration when x=3x=3.
Part a · 2 marks

Find the acceleration in metres per second squared.

Worked solution

Answer: -77/125

  1. Differentiate Velocity: difv/difx=−7/(2x−1)2dif v/dif x=-7/(2x-1)^2.
  2. Acceleration Chain Rule: -77/125

Catalog reference: ext2_jr25_acceleration_from_position_velocity · seed 41001

Worked example 2

1 markext2-mechanics-1
Maximum acceleration in SHM A particle moves in simple harmonic motion with v2=2304−48x−2x2v^2=2304-48x-2x^2. What is the maximum magnitude of its acceleration?
  1. A.24 24 metres per second squared.
  2. B.48 48 metres per second squared.
  3. C.72 72 metres per second squared.
  4. D.96 96 metres per second squared.
Part a · 1 mark

Select the acceleration in metres per second squared.

Worked solution

Answer: 72

  1. Solve Quadratic: The turning positions, where v=0v=0, are x=8sx=8{{s}} and x=−16sx=-16{{s}}.
  2. Differentiate Velocity Relation: a=−8s−2xa=-8{{s}}-2x, so the larger endpoint magnitude is 24s24{{s}}.

Catalog reference: ext2_cssa25_shm_maximum_acceleration · seed 41001

Worked example 3

1 markext2-mechanics-1
Distance in simple harmonic motion A particle satisfies v2=(pi2/9)(42−x2)v^2=(pi^2/9)(4^2-x^2). Find the distance travelled in the first 60 seconds.
  1. A.80
  2. B.96
  3. C.160
  4. D.192
Part a · 1 mark

Select the distance in metres.

Worked solution

Answer: 160

  1. Shm Period: 6
  2. Complete Cycles: 160

Catalog reference: ext2_ind25_mcq_shm_minute_distance · seed 41001

Content mapping and common errors

Mechanics content identifiers

MEX-M1

Applications of Calculus to Mechanics

Model rectilinear and projectile motion; work with simple harmonic motion; analyse resisted motion.

Common student mistakes
  • ·Confusing displacement, velocity and acceleration (especially signs)
  • ·SHM: forgetting that a = −n²x requires the restoring force towards the centre
  • ·Incorrect integration of equations of motion under resistance

Source and methodology

Examples are deterministic curriq catalog generations from existing QDSL archetypes at the published fixed seed. The generator copies only student-facing prompts, mark allocations, topic mappings, display tables or options, checked answers and marking-path results into this resource.

Review the official NSW Curriculum source ↗

Other Extension 2 topics

Proof
2 content identifiers
Vectors
1 content identifier
Complex Numbers
2 content identifiers
Calculus
2 content identifiers

From the blog

Formula sheet
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HSC Maths Past Papers by Topic

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