AQA A-Level Physics Paper 1, June 2022: Question 23

1 mark · Medium difficulty · Multiple Choice

Calculate the total distance travelled by a ball thrown vertically upwards that returns to its original position after 2.4 s.

Practise this question

Question

Multiple choice question 23 asks for the total distance travelled by a ball thrown vertically upwards that returns to its original position 2.4 s later, with negligible air resistance. Four options are given: A 5.9 m, B 7.1 m, C 14 m, and D 28 m.
Question text

23 A ball is thrown vertically upwards and returns to its original position 2.4 s later.

The effect of air resistance is negligible.

What is the total distance travelled by the ball?

[1 mark]

A 5.9 m

B 7.1 m

C 14 m

D 28 m

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer is option C (14 m).

23 C (AO2) 14 m

How to answer it

Vertical Projectile Motion: Total Distance

What this question tests

This question assesses your understanding of constant acceleration equations of motion (suvat) under gravity, the symmetry of vertical projectile motion, and the critical distinction between displacement and total distance travelled.

Question 23

Exam Breakdown & Solution

✅ Correct Answer

C (14 m)

The correct option is C. The total distance travelled by the ball up and down combined is 14 metres.

💡 Key Knowledge

  • Free Fall Acceleration: Acceleration g = 9.81 m s⁻² (or 9.8 m s⁻² ) downwards.
  • Symmetry of Flight: The time taken to reach maximum height is exactly half of the total flight time.
  • Distance vs. Displacement: Since displacement returns to zero when the ball hits its starting point, you must calculate the maximum height reached in one direction and double it.

🧠 Exam Technique

  • Split the motion into manageable halves: consider the upward journey only, taking time t = 1.2 s .
  • At the highest point, final velocity v = 0 m s⁻¹ .
  • Use a suvat equation that does not require initial velocity, such as s = vt - (1/2)at² applied backwards, or find initial velocity first using v = u + at .

❌ Common Errors

  • Displacement Confusion: Confusing total distance with total displacement (which would incorrectly yield 0 m ).
  • Time Halving Error: Using the full time 2.4 s in suvat equations while treating final velocity as zero at that instant.
  • Forgetting to Double: Calculating only the maximum height (approx. 7.1 m , Option B) and failing to account for the downward trip back to the hand.

📐 Step-by-Step Calculation

  1. Find the time to peak ( t ):
    Total time = 2.4 s . Time to maximum height t = 2.4 / 2 = 1.2 s .
  2. Identify known variables for the upward journey:
    a = -9.81 m s⁻² (taking upwards as positive)
    t = 1.2 s
    v = 0 m s⁻¹ (at maximum height)
  3. Calculate maximum height ( s ) using suvat:
    Using s = vt - (1/2)at² (considering motion from peak to start, or standard suvat):
    First find initial velocity u using v = u + at → 0 = u + (-9.81 × 1.2) → u = 11.772 m s⁻¹ .
    Then use s = ut + (1/2)at² :
    s = (11.772 × 1.2) + 0.5 × (-9.81) × (1.2)² = 14.1264 - 7.0632 = 7.0632 m .
  4. Calculate total distance:
    Total distance = 2 × s = 2 × 7.0632 = 14.1264 m , which rounds to 14 m (matching Option C).
Examiner Note: Top-level candidates instantly recognised that time to peak is half the total time and correctly doubled the single-direction displacement. Option B ( 7.1 m ) was a very common distractor chosen by students who calculated only the height of the upward journey and forgot the return trip.

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.