AQA GCSE Physics Physics Paper 2 (Higher), June 2022: Question 4
12 marks · Standard Demand difficulty · Short Answer
Explain the forces acting on falling hailstones reaching terminal velocity, compare their kinetic energy, and calculate the average stopping force on impact.
Practise this questionQuestion
Question text
04 Hailstones are small balls of ice. Hailstones form in clouds and fall to the ground.
Figure 7 shows different-sized hailstones.
Figure 7
A hailstone falls from a cloud and accelerates.
04.1 Why does the hailstone accelerate?
[1 mark]
04.2 The hailstone stops accelerating and reaches terminal velocity.
Explain why the hailstone reaches terminal velocity.
[3 marks]
A scientist investigated how the mass of hailstones affects their terminal velocity.
Figure 8 shows the results.
Figure 8
04.3 Why does terminal velocity increase with mass?
[1 mark]
Tick ( ) one box.
As mass increases the cross-sectional surface area of a hailstone
increases.
As mass increases the volume of a hailstone increases.
As mass increases the weight of a hailstone increases.17
04.4 Explain the difference in the maximum kinetic energy of a hailstone with a mass of
10 g and a hailstone with a mass of 20 g.
[3 marks]
04.5 The kinetic energy of a hailstone is measured in joules.
Which of the following is the same as 1 joule?
[1 mark]
Tick ( ) one box.
1 Nm
1 N/m
1 N/m2
1 N m2 18
Figure 8 is repeated below.
Figure 8
04.6 A hailstone hit the ground at its terminal velocity of 25 m/s.
The hailstone took 0.060 s to stop moving.
Determine the average force on the hailstone as it hit the ground.
Use information from Figure 8.
Use the Physics Equations Sheet.
[3 marks]
Average force = N
Mark scheme
Show the mark scheme
Question 4
AO /
Question Answers Extra information Mark
Spec. Ref.
04.1 there is a resultant force acting allow weight/gravity is greater 1 AO1
than air resistance 4.5.6.1.5
allow (initially) weight/gravity is
the only force acting
AO /
Spec. Ref.
04.2 as the velocity of the hailstone allow speed for velocity 1 AO1
increases air resistance 4.5.6.1.5
increases
until air resistance becomes 1
equal to the weight of the
hailstone
so the resultant force is (equal 1
to) zero
AO /
Spec. Ref.
04.3 as mass increases the weight of 1 AO3
a hailstone increases 4.5.6.1.5
AO /
Spec. Ref.
04.4 kinetic energy depends on both allow E = ½ mv2 1 AO1
k
mass and velocity
as mass increases so does a statement is required 1 AO1
terminal / maximum velocity
kinetic energy ∝ m and kinetic this mark can be scored by 1 AO3
energy ∝ v2 so as mass doubles relevant calculations
kinetic energy more than 4.1.1.2 15
doubles
AO /
Spec. Ref.
04.5 1 N m 1 AO3
4.5.2
AO /
Spec. Ref.
04.6 mass = 0.0185 (kg) allow 0.018 to 0.019 inclusive 1 AO2
4.5.7.3
0.0185 × 25 allow a correct substitution using 1
F = an incorrectly / not converted
0.060
value of m
F = 7.708 (N) allow 7.7 (N) 1
allow correct calculation using
an incorrectly / not converted
value of m
if no other marks are awarded
a misreading of the scale giving
a value between 15.6 and 15.7
inclusive that is then correctly
converted giving an answer
between 6.50 and 6.54 scores 2
marks
a misreading of the scale giving
a value between 15.6 and 15.7
inclusive that is then not
16 converted giving an answer
between 6500 and 6542 scores
1 mark
Total Question 4 12
How to answer it
Terminal Velocity, Kinetic Energy & Impact Forces
This question assesses key concepts from Forces and Motion and Energy:
- Applying Newton’s laws to falling objects and explaining how terminal velocity is reached.
- Understanding how weight and mass affect drag requirements and final speed.
- Evaluating the relationship between mass, velocity, and kinetic energy (Ek = ½mv²).
- Recognising derived equivalent units (Joules and Newton-metres).
- Extracting data from a non-linear graph, converting units (g to kg), and calculating average impact force (F = mΔv / Δt).
Question 04.1
Resultant Force and Initial Acceleration [1 Mark]
✅ Correct Answer
Any one of the following:
- There is a resultant force acting on the hailstone.
- Weight (gravity) is greater than air resistance.
- Initially, weight is the only force acting.
💡 Key Knowledge
Newton's First & Second Laws: An object will only accelerate if there is an unbalanced (resultant) force acting on it ( F = ma ). At the start of a fall, speed is zero so drag is zero; weight acts downwards alone.
Question 04.2
Explaining Terminal Velocity [3 Marks]
✅ 3-Step Model Answer
- As the hailstone’s velocity increases, air resistance increases. [1 mark]
- Air resistance increases until it becomes equal to the weight of the hailstone. [1 mark]
- The resultant force becomes zero (forces are balanced), so acceleration stops. [1 mark]
🧠 Exam Technique
Always structure terminal velocity answers as a 3-part timeline:
Speed up → Drag increases → Drag equals Weight (Resultant Force = 0).
Missing out the phrase "resultant force is zero" is the #1 reason students drop the final mark!
❌ Common Errors
- Claiming that gravity "decreases" or "runs out" as it falls (weight remains constant!).
- Saying "air resistance becomes greater than weight" (that would cause it to decelerate, not fall at constant speed).
Question 04.3
Mass vs. Terminal Velocity [1 Mark]
✅ Correct Choice
☑ As mass increases the weight of a hailstone increases.
💡 Why is this correct?
Terminal velocity is reached when drag = weight. Since W = mg , a greater mass means a greater downward weight force. Therefore, the hailstone must fall much faster to generate a large enough air resistance to balance that larger weight.
Question 04.4
Comparing Kinetic Energy: 10 g vs. 20 g Hailstones [3 Marks]
✅ Model Answer
- Kinetic energy depends on both mass and velocity ( Ek = ½mv² ). [1 mark]
- As mass increases, terminal velocity also increases. [1 mark]
- Because Ek ∝ m and Ek ∝ v² , doubling the mass means the kinetic energy more than doubles. [1 mark]
📐 Mathematical Proof (Alternative Route)
You can also score full marks using graph values:
- At 10 g: v ≈ 5 m/s
Ek = ½ × 0.010 × 5² = 0.125 J - At 20 g: v ≈ 32 m/s
Ek = ½ × 0.020 × 32² = 10.24 J - 10.24 J ≫ 0.125 J , clearly showing kinetic energy increases massively (by ~80×!).
❌ Common Mistake
Assuming that doubling the mass simply doubles the kinetic energy. This overlooks that velocity also increases, and kinetic energy is proportional to the square of velocity ( v² ).
Question 04.5
Unit of Energy: Equivalent to 1 Joule [1 Mark]
✅ Correct Choice
☑ 1 N m (Newton-metre)
💡 Formula Link
Work Done = Force × Distance ( W = F × s )
Units: Joules (J) = Newtons (N) × metres (m) = N m
Question 04.6
Calculating Average Impact Force [3 Marks]
📐 Step-by-Step Calculation
Step 1: Read the mass from Figure 8 at v = 25 m/s
Find 25 m/s on the y-axis, trace horizontally to the curve, then down to the x-axis:
Mass = 18.5 g (Mark scheme allows 18 to 19 g).
Step 2: Convert mass to standard SI units (kg)
m = 18.5 ÷ 1000 = 0.0185 kg
[1 mark] awarded for converting mass into kg.
Step 3: Choose equation and substitute values
From the Physics Equations Sheet, Force equals rate of change of momentum (or F = ma where a = Δv / t ):
F = (m × Δv) ÷ t
F = (0.0185 kg × 25 m/s) ÷ 0.060 s
[1 mark] for correct substitution.
Step 4: Calculate final answer
F = 0.4625 ÷ 0.060 = 7.708 N (or 7.7 N)
Average Force = 7.7 N (or 7.71 N) [1 mark]
❌ Calculation Traps
- Forgetting to convert grams to kilograms: Using 18.5 instead of 0.0185 gives an answer of 7708 N (loses 1 mark).
- Misreading the scale: On the mass axis, 10 small squares = 5 g, so 1 small square = 0.5 g.
🧠 Graph Reading Check
Between 15 g and 20 g there are 10 small grid squares. At 25 m/s, the curve intersects at 7 small squares past 15:
15 + (7 × 0.5) = 18.5 g
Topics
Physics · P1: Energy · P5: Forces
Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Higher), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.