AQA AS Level Physics Paper 1, June 2024: Question 6
12 marks · Hard difficulty · Extended Answer
Analyze energy changes, forces, and motion of a bungee jumper using graphs and equations for gravitational potential energy, kinetic energy, Hooke's law, and the Young modulus.
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Question text
06 Figure 16 shows a boy of mass m standing on a platform about to perform a bungee
jump. He steps off the platform and falls vertically. The tension in the rope increases
as it stretches. The boy decelerates to rest at the lowest point of the jump.
Assume that air resistance is negligible throughout this question.
Figure 16
During the jump, s is the vertical displacement moved by the boy’s centre of mass.
The lowest point of the jump occurs when s is 27 m.
Figure 17 shows the variation of his velocity v with s during the jump.
Figure 17
06.1 The boy experiences freefall when he steps off the platform.
During which part of the jump does the boy’s acceleration begin to decrease?
Tick ( ) one box.
*22* [1 mark]
between s = 0 and s = 7.5 m
between s = 7.5 m and s = 15 m
between s = 15 m and s = 22.5 m
between s = 22.5 m and s = 27 m
06.2 When the boy’s centre of mass has moved through a distance s of 15.0 m the change
in his gravitational potential energy is 9.56 kJ.
Calculate the mass m of the boy.
[2 marks]
24 m = kg
The bungee rope has a stiffness k of 110 N m−1 and obeys Hooke’s law.
06.3 The maximum kinetic energy of the boy is 7.71 kJ.
Calculate, by considering the energy transfers, the extension ΔL of the bungee rope
when the kinetic energy of the boy is at a maximum.
[3 marks]
ΔL = m
06.4 Deduce the tension in the rope when the kinetic energy of the boy is at a maximum.
Give a reason to support your answer.
[2 marks]
tension = N
reason
The original rope is replaced with a second rope and the boy repeats the jump.
Table 1 contains information about the original rope and the second rope. Both ropes
*24* obey Hooke’s law.
Table 1
Young modulus Cross-sectional area Unstretched length
original rope E A L
second rope 1.2E A 1.2L
The Young modulus is given by:
stiffness×unstretched length
Youngmodulus =
cross - sectional area
06.5 Show that each rope has the same stiffness.
[1 mark]
06.6 Deduce whether the boy’s maximum velocity is increased when using the
second rope.
[3 marks]
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark AO
06.1 between s = 7.5 m and s = 15 m Tick in 2nd box only 1 AO3
06.2 Use of ΔEP = mgh Use of: rearrangement where m would be 2 1 × AO1
subject or substitution. 1 × AO2
Condone one error in substitution.
Calculator display =
(m = )65(.0) (kg) For g = 9.81 ms-2 = 64.96772001
For g = 9.8 ms-2 = 65.0340136054421
Alternative method for an ECF from 06.1 (tick
in 3rd or 4th boxes).
• Use of 𝐸𝐸𝑘𝑘 = 𝑚𝑚𝑑𝑑
OR
Read-off for v = 15.4 ms-1
(Acceptable range 15.2 ms-1 to
15.6 ms-1)
• m= 80.6 (kg)
(Acceptable range 78.57 kg to 82.76 kg)
06.3 Max 2 from: Accept correct energy conservation statement 3 1 × AO1
E = 9.56 – 7.71 = 1.85 (kJ) for MP1 2 × AO2
• Energy difference ( )
For example:
∆𝐸𝐸𝑃𝑃 = 𝐸𝐸𝐾𝐾 + 𝑒𝑒𝑑𝑑𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑠𝑠𝑡𝑡𝑑𝑑𝑒𝑒𝑒𝑒𝑑𝑑 (𝑒𝑒𝑑𝑑 𝑒𝑒𝑑𝑑𝑟𝑟𝑒𝑒)
Use of:
• Use of E = k∆L Rearrangement to make ∆𝐿𝐿 the subject or by
2 substitution.
Condone use of their 𝐸𝐸 and one other error in
substitution. (allow 9.56 (kJ) or 7.71(kJ) for E)
Condone use of
E = F∆L and F = k∆L OR
E = F∆L and F=mg
With their F and their E seen in E = F∆L
Must be an energy difference. Condone POT
• ΔL = �2 ×𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑒𝑒𝑎𝑎𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑑𝑑𝑒𝑒𝑓𝑓𝑓𝑓𝑒𝑒𝑒𝑒𝑒𝑒𝑎𝑎𝑐𝑐𝑒𝑒 9.56 (kJ) 7.71(kJ)
Do not accept or for their
𝑘𝑘
energy difference.
Max 1 mark for:
637.65 = 110 × ∆𝐿𝐿 giving ΔL = 5.8 m
must be done by considering energy transfers.
ΔL = 5.8(0) m OR
answer without working.
06.4 (Tension =) 640 (N) Potential ECF from: 2 1 × AO2
• m in 06.2 where use T=mg 1 × AO3
• ∆𝐿𝐿 in 06.3 (typical ecf answer =1300
(N) where use T=k∆𝐿𝐿
Reason:
Idea that the resultant force / acceleration is upwards (in
opposite direction to motion) for tension greater than this For two marks:
value. Reason must be consistent with any working
OR seen.
Idea that the resultant force / acceleration is downwards (in
same direction as motion) for tension less than this value Insufficient to state that tension = weight at
maximum kinetic energy.
OR
Resultant force / acceleration is zero (when kinetic energy is
at its maximum.) Apply list rules to the reason.
OR
Tension is directly proportional to the extension / (rope obeys)
If use F=k ∆𝐿𝐿 without further support in their
Hooke’s law.
reason can score max 1 mark.
e.g. Each term to be defined
06.5 EA Accept 1 AO1
Use of k = to show k is same for both ropes 𝑘𝑘 ×1.2𝐿𝐿 𝑘𝑘 ×1.2 𝐿𝐿 𝑘𝑘 × 𝐿𝐿
L 1.2 E = ⇒ 1.2 E = 𝐴𝐴 ⇒ E =
𝐴𝐴 𝐴𝐴
30 Or equivalent
Allow use of k =110 Nm-1 in working.
06.6 Yes: Must have correct deduction for 3 marks. 3 3 × AO3
MAX 2 from:
• (Second) rope’s (unstretched) length is greater.
• Has a greater velocity before rope begins to stretch
(for second rope).
• Extension of each rope is same (when tension =
weight.)
• Work done in stretching rope is same (in travelling to
max velocity) / energy stored in rope is same
• Total distance fallen to reach max velocity is greater
(for second rope)
• Total distance fallen (to max velocity) = unstretched
length + same extension
• Idea of longer time in free-fall
Conservation of energy:
Gains more kinetic energy before work done
Correct use of principle of conservation of energy or correct by tension becomes greater than work done by
use of Newton’s 2nd law gravity. 31
Newton’s 2nd law:
Gains more velocity before acceleration’s
direction becomes opposite to motion’s
direction.
Total 12
How to answer it
Bungee Jump Mechanics, Energy Transfers and Material Properties
What this question tests
This question assesses core mechanics and energy concepts applied to a vertical bungee jump. Key skills include interpreting velocity-displacement graphs, applying the principle of conservation of energy (gravitational potential energy, kinetic energy, and elastic strain energy), analyzing forces using Newton's second law, and evaluating how changes to a material's dimensions and Young modulus affect system dynamics.
Acceleration and Forces in Freefall
✅ Correct Answer
between s = 7.5 m and s = 15 m (Tick the 2nd box only)
💡 Key Knowledge
The boy experiences freefall (or negligible tension) until the rope begins to stretch. From the velocity-displacement graph, the maximum velocity (where acceleration is zero, meaning weight equals tension) occurs at s = 15 m . Acceleration begins to decrease as soon as the upward tension force starts opposing gravity, which happens once the rope becomes taut ( s > 7.5 m ).
Calculating the Mass of the Boy
📐 Step-by-Step Calculation
- Identify that loss in gravitational potential energy equals gain in kinetic energy before the rope stretches: ΔEₚ = Eₖ
- Substitute formulas: mgΔs = 0.5 × m × v² (mass m cancels out, meaning velocity at a given height is independent of mass, or use data from the graph at s = 15.0 m where ΔEₚ = 9.56 kJ = 9560 J ).
- Rearrange for mass: m = ΔEₚ / (g × Δs) or m = 9560 / (9.81 × 15.0)
- Calculate: m = 65.0 kg (using g = 9.81 m s⁻² gives ~64.97 kg; using g = 9.8 m s⁻² gives ~65.03 kg).
❌ Common Errors & Exam Technique
Students often forget to convert kiloJoules to Joules ( 9.56 kJ = 9560 J ). Ensure powers of 10 are handled correctly. Condone one substitution error if rearrangement is correct.
Rope Extension at Maximum Kinetic Energy
📐 Step-by-Step Calculation
- Find the energy stored in the rope by considering conservation of energy: Energy difference = ΔEₚ - Eₖ(max)
- Calculate stored elastic strain energy: E = 9.56 kJ - 7.71 kJ = 1.85 kJ = 1850 J
- Equate to elastic potential energy formula: E = 0.5 × k × (ΔL)²
- Rearrange for extension ΔL : ΔL = sqrt((2 × E) / k)
- Substitute values ( k = 110 N m⁻¹ ): ΔL = sqrt((2 × 1850) / 110) = sqrt(33.64) = 5.80 m
🧠 Exam Technique
State your energy balance clearly at the start. Examiners look for explicit recognition that the lost gravitational potential energy goes partly into kinetic energy and partly into elastic strain energy stored in the stretched rope.
Tension at Maximum Kinetic Energy
✅ Correct Answer
Tension = 640 N
Reason: At maximum kinetic energy, acceleration is zero, so the upward tension equals the downward gravitational force (weight, W = mg = 65.0 kg × 9.81 m s⁻² = 638 N ≈ 640 N ). Alternatively, can use T = kΔL = 110 × 5.80 = 638 N .
❌ Common Errors
A common misconception is thinking tension is zero or maximum at peak kinetic energy. Remember: maximum kinetic energy occurs where net force is zero ( T = W ).
Comparing Stiffness of Both Ropes
💡 Key Knowledge & Proof
From the definition given: Young modulus E = (stiffness k × unstretched length L) / cross-sectional area A .
Rearranging for stiffness k : k = (EA) / L
For the second rope: k₂ = ((1.2E) × A) / (1.2L) = (EA) / L = k₁ . Thus, both ropes have identical stiffness.
Maximum Velocity with the Second Rope
✅ Correct Answer
Yes (Maximum velocity is increased).
💡 Examiner Commentary & Key Points
- The second rope has a greater unstretched length ( 1.2L ) while maintaining the same stiffness ( k ).
- Because the unstretched length is greater, the boy falls a greater distance before the rope begins to stretch and slow him down.
- More gravitational potential energy is converted into kinetic energy before deceleration begins, leading to a higher maximum velocity.
Topics
Physics · Practical skills · Required Practicals · 3.4 Mechanics and materials · Data analysis · AS practicals (1–6)
Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.