AQA GCSE Physics Physics Paper 2 (Higher), June 2025: Question 5

💡 Diagram Requirements

• Starting point: The arrow must start directly from the central dot representing the skydiver's centre of mass.
• Direction: Vertically upwards (directly opposite to weight).
• Relative length: Must be noticeably shorter than the downward weight arrow because the skydiver is still accelerating downwards.
• Label: Labelled clearly as air resistance or drag.

❌ Common Errors

• Drawing the upward arrow equal in length to weight (that would represent terminal velocity!).
• Drawing an arrow longer than weight (which would mean the skydiver is decelerating).
• Labeling the upward force as "upthrust" (upthrust is the buoyancy force in fluids, not the drag resisting fast motion).
• Floating arrows not attached to the central dot.

📐 Step-by-Step Calculation

1. Convert time to SI units (seconds):
   t = 2.5 min = 2.5 × 60 = 150 s [1 mark]
2. Select and substitute into formula:
   a = (v − u) / t
   0.64 = (v − 0) / 150 [1 mark]
3. Rearrange to solve for final velocity (v):
   v = 0.64 × 150 [1 mark]
4. State the answer:
   v = 96 m/s [1 mark]

🧠 Exam Technique & Pitfalls

• Unit trap: Never calculate with time in minutes when acceleration is in m/s² . Always convert minutes to seconds first!
• Initial velocity: The question states "initial velocity was 0 m/s", so u = 0 .
• Alternative formula form: v = u + at = 0 + (0.64 × 150) = 96 m/s is equally valid and directly gives v .

✅ Model Answer (Point-by-Point)

1. As the skydiver’s speed / velocity increases, the air resistance increases. [1]
2. Air resistance increases until it becomes equal in magnitude to the weight of the skydiver (acting in the opposite direction). [1]
3. Therefore, the resultant force becomes zero. [1]
4. According to Newton’s First Law, when resultant force is zero, acceleration becomes zero and the skydiver falls at a constant velocity. [1]

🧠 Examiner Commentary

• Cause & Effect: Top-scoring responses follow the sequence: Speed increases → Drag increases → Balanced forces (resultant force = 0) → Acceleration = 0 (terminal velocity reached).
• Weight remains constant: Do not state that "weight decreases". The downward pull of gravity remains constant; it is the drag that changes!

✅ Two Accepted Explanations

Approach 1 (Weight of air above):

• As the skydiver falls, there is a greater weight (or mass/number) of air molecules above them. [1]
• Since pressure is force per unit area ( P = F / A ), this greater weight causes pressure to increase. [1]

Approach 2 (Air density and collisions):

• The density of air increases closer to the Earth's surface. [1]
• This causes more frequent collisions between air particles and the skydiver. [1]

❌ Common Errors

• Saying simply "the air gets heavier" without linking it to the volume/column of air above the skydiver.
• Forgetting to link particle density to collision frequency: simply saying "there are more particles" is insufficient for the second mark.

📐 How to Construct the Vector Diagram

1. Choose a scale: E.g., 1 cm = 40 N (or 1 large grid square = 40 N or 50 N).
   • Vertical force = 240 N upwards → 6 cm line straight up.
   • Horizontal force = 200 N to the left → 5 cm line to the left.
2. Draw tip-to-tail triangle or parallelogram:
   Ensure the vertical side is visibly longer than the horizontal side (240 N > 200 N). [1 mark]
3. Draw resultant force vector:
   An arrow starting from the origin and pointing diagonally upwards and to the left (the hypotenuse). [1 mark]
4. Measure magnitude and angle:
   Measure length with a ruler and convert via scale. Measure angle from vertical with a protractor.

✅ Final Answers & Acceptable Ranges

• Magnitude:
  Theoretical value: √(240² + 200²) = √(57600 + 40000) = 312.4 N
  Acceptable mark range: 300 N to 320 N [1 mark]
• Angle to the vertical:
  Theoretical value: tan(θ) = 200 / 240 ⇒ θ = 39.8°
  Acceptable mark range: 38° to 42° [1 mark]