AQA A-Level Physics Paper 1, June 2024: Question 21

1 mark · Medium difficulty · Multiple Choice

Calculate the horizontal distance travelled by a particle with a constant horizontal velocity and a constant vertical acceleration given its vertical displacement.

Practise this question

Question

Multiple choice question 21 asking for the horizontal distance a particle travels given its horizontal speed of 1.0 x 10^7 m s^-1, constant vertical acceleration of 4 x 10^14 m s^-2, and vertical displacement of 8 x 10^-2 m. Four options are provided: A 0.2 m, B 0.1 m, C 2 x 10^-8 m, and D 0.4 x 10^-9 m.
Question text

21 A particle travelling horizontally at 1.0 × 107 m s−1 enters a region where it has a constant

vertical acceleration of 4 × 1014 m s−2.

What is the horizontal distance the particle has travelled in the region when its vertical

displacement is 8 × 10−2 m?

[1 mark]

A 0.2 m

B 0.1 m

C 2 × 10−8 m

D 0.4 × 10−9 m

Mark scheme

Show the mark scheme Mark scheme table indicating question number 21 has the correct answer A, corresponding to 0.2 m, assessed under AO2.

21 A 0.2 m AO2

How to answer it

Particle Motion & Projectile Kinematics

What this question tests

This question assesses your ability to apply SUVAT kinematic equations to two-dimensional motion (specifically separating horizontal and vertical components), handling constant acceleration, and calculating horizontal distance travelled using time as the linking variable.

Question 21 (Multiple Choice) — AO2 (Application)

Solution & Breakdown

✅ Correct Answer

A ( 0.2 m )

The correct option is A. Let's look at how to arrive at this result step-by-step.

💡 Key Knowledge

  • Horizontal and vertical motions are independent of one another.
  • Horizontally: velocity is constant ( u = v ), so s = v × t .
  • Vertically: constant acceleration applies, so use SUVAT: s = u₀t + ½at² . Assuming initial vertical velocity u₀ = 0 , this simplifies to s = ½at² .

🧠 Exam Technique

For multi-step multiple choice questions involving kinematics:

  1. Identify the variable that connects both directions: Time ( t ).
  2. Calculate the time using the vertical motion data first.
  3. Substitute that exact time into the horizontal motion equation. Do not prematurely round intermediate values!

❌ Common Errors

  • Mixing up axes: Mistakenly applying the vertical acceleration value into the horizontal distance equation ( s = vt ).
  • Power of 10 slips: Forgetting to handle standard form indices properly (e.g., miscalculating square roots of negative or fractional exponents).

📐 Step-by-Step Calculation

  1. Find the time ( t ) from vertical motion:
    Using s = ½at² (with initial vertical velocity u = 0 ):
    8 × 10⁻² = ½ × (4 × 10¹⁴) × t²
    8 × 10⁻² = (2 × 10¹⁴) × t²
    t² = (8 × 10⁻²) / (2 × 10¹⁴) = 4 × 10⁻¹⁶ s²
    t = √(4 × 10⁻¹⁶) = 2 × 10⁻⁸ s
  2. Calculate horizontal distance ( s ):
    Using s = v × t for constant horizontal velocity:
    s = (1.0 × 10⁷ m s⁻¹) × (2 × 10⁻⁸ s)
    s = 0.2 m

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.