AQA AS Level Physics Paper 2, June 2025: Question 20

1 mark · Easy difficulty · Multiple Choice

Determine the proportionality relationship between the velocity of a freely falling object from rest and the distance fallen.

Practise this question

Question

Multiple choice question asking: An object falls from rest. Air resistance is negligible. When the object has fallen through a distance x, its velocity is v. How does v vary with x? Options are: A: v proportional to square root of x; B: v proportional to 1/x; C: v proportional to x squared; D: v proportional to 1/(x squared).

Mark scheme

Show the mark scheme Mark scheme table row for question 20 showing the correct answer as option A, with reference to AO2 and v is proportional to the square root of x.

How to answer it

Relationship Between Velocity and Distance in Free Fall

What this question tests

This question tests your ability to apply the equations of uniformly accelerated motion (SUVAT) or the principle of conservation of energy to derive a proportionality relationship between final velocity and displacement for an object falling freely under gravity.

Question 20 (Multiple Choice)

Identifying how velocity (v) varies with distance fallen (x) from rest

✅ Correct Answer

A: v ∝ √x

Mark Award: [1 Mark] (Assessment Objective AO2 - Application of knowledge in a theoretical context).

💡 Key Knowledge

  • "From rest": Initial velocity u = 0 m s⁻¹ .
  • Negligible air resistance: Acceleration is constant and equal to the acceleration due to gravity, a = g = 9.81 m s⁻² .
  • SUVAT Equation: Connects initial speed ( u ), final speed ( v ), acceleration ( a ), and displacement ( s = x ):
    v² = u² + 2as
  • Alternative (Energy): Loss of Ep = Gain in Ek :
    m g x = ½ m v²

📐 Step-by-Step Derivation

Method 1: Using Kinematics (SUVAT)

  1. Identify variables: u = 0 , a = g , s = x .
  2. Substitute into v² = u² + 2as :
    v² = 0² + 2gx = 2gx
  3. Take the square root of both sides:
    v = √(2gx) = √(2g) · √x
  4. Since 2g is a constant, v ∝ √x .

Method 2: Using Conservation of Energy

  1. Equate kinetic and potential energy:
    ½mv² = mgx
  2. Divide both sides by mass m and rearrange:
    v² = 2gx ⇒ v ∝ √x

🧠 Exam Technique & Elimination

  • Look for physical reality first: As an object falls, it speeds up, meaning v must increase as x increases.
  • Instantly eliminate B and D: Both v ∝ 1/x and v ∝ 1/x² represent inverse relationships (speed decreasing as distance increases), which is physically impossible here.
  • Choose between A and C: Recall that displacement increases with the square of time ( x ∝ t² ) while velocity increases linearly with time ( v ∝ t ). Therefore, substituting gives x ∝ v² , which rearranges directly to v ∝ √x .

❌ Common Errors & Examiner Traps

  • Confusing the subject of proportionality: Remembering x ∝ v² but carelessly picking option C ( v ∝ x² ) instead of taking the square root.
  • Confusing distance with time: Remembering that v = gt (so v ∝ t ) and mistakenly assuming that v is also directly proportional or square proportional to x without checking the equations.
  • Misreading the question: Forgetting that u = 0 ("from rest"), which simplifies the equation directly to a pure monomial proportionality.

Topics

Physics · 3.4 Mechanics and materials

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