AQA A-Level Physics Paper 1, November 2020: Question 21

Falling Water Drops Motion Analysis

AQA A-Level Physics • Mechanics • Motion in One Dimension

What this question tests

This question assesses your understanding of constant acceleration equations (suvat) applied to multiple moving objects under gravity. It tests your ability to reason about relative time intervals, free-fall kinematics, and distance-time relationships without getting bogged down in unnecessary full calculations.

Question 21

Multiple Choice Component [1 mark]

✅ Correct Answer: B (0.20 m)

When the lead drop hits the surface, the previous drop has been falling for a shorter time interval and is positioned at a height corresponding to option B.

💡 Key Knowledge

Drops leave at regular time intervals (delta t = 0.20 s).
All drops experience the same constant acceleration due to gravity ( g ).
The total fall distance s = 0.80 m allows us to find the total fall time for any drop.

🧠 Exam Technique

Avoid calculating absolute times if you can use proportional reasoning. Notice that distance is proportional to time squared ( s ∝ t² ) from rest, which lets you compare relative positions efficiently under time scaling.

❌ Common Errors

Subtracting 0.20 s directly from the total distance (assuming linear speed/distance relationships).
Confusing the time the drop has been falling with the time interval between drops.

📐 Step-by-Step Calculation & Logic

Find total fall time ( T ) for the drop hitting the surface:
Using s = ut + ½gt² with u = 0 :
0.80 = ½ × 9.81 × T² ⇒ T ≈ 0.404 s (or use working with g as 9.8 or 9.81 ).
Determine time fallen by the previous drop ( t ):
Since drops leave at intervals of 0.20 s, the previous drop started 0.20 s earlier, meaning when the first drop hits, the second drop has been falling for:
t = T - 0.20 s = 0.404 - 0.20 = 0.204 s (exactly half of the total time T ).
Calculate distance fallen in that time ( s_2 ):
s_2 = ½ g t²
Since time t is half of T , squaring it means t² = (½ T)² = ¼ T² .
Therefore, the distance fallen is ¼ of the total distance:
s_2 = ¼ × 0.80 m = 0.20 m .

Examiner Note: This is a classic A-Level multiple-choice discriminator. Top-performing students immediately recognized the s ∝ t² scaling factor, saving valuable minutes in the exam.