AQA AS Level Physics Paper 1, June 2022: Question 7
13 marks · Medium difficulty · Extended Answer
Analyze the mechanics and energy transfers of a cyclist moving up and down a hill, including force components, power output, velocity-time graphs, and acceleration.
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Question text
07.1 Figure 8 shows a cyclist going up a hill.
Figure 8
The angle θ of the slope of the hill is constant.
The total mass m of the cyclist and bicycle is 65 kg.
Write an expression for the component of the total weight parallel to the slope.
[1 mark]
07.2 The useful power output of the cyclist is 310 W.
The cyclist has a steady speed of 1.63 m s−1.
Assume that air resistance is negligible at this speed.
Calculate θ.
[2 marks]
θ = °
Figure 9 shows an alternative ‘zig-zag’ path taken by the cyclist up the same hill.
She maintains a steady speed of 1.63 m s−1.
Figure 9
07.3 Discuss how her useful power output when taking the path in Figure 9 compares with
her useful power output in Question 07.2.
[3 marks]
The cyclist reaches the top of the hill. She then travels back down the hill in a straight
line. The bicycle rolls freely without the cyclist pushing the pedals or applying the
brakes.
Figure 10 shows the variation of her velocity with time as she goes down the hill.
Figure 10
07.4 Determine the acceleration of the cyclist 10.0 s after she begins to go down the hill.
[3 marks]
19acceleration = m s−2
07.5 Energy transfers occur as the cyclist travels down the hill.
Outline how these energy transfers explain the shape of the graph in Figure 10.
*18* [4 marks]
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark AO
07.1 Allow mg sin θ or 65g sin θ or 638 sin θ 1 AO2
or 637.7 sin θ or 637.65 sin θ
Condone labelling this component as W
(Component of total weight parallel to slope =) 640 sin θ in statements such as
W= 640 sin θ
Do not accept
W sin θ unless W is defined as mg
07.2 use of P = Fv Ecf from 07.1 for MP1 and MP2 2 1 x AO1
Use of P = Fv by substitution and rearrangement to 1 x AO2
make F the subject.
Expect to see (F =) 190(.184) (N)
Accept a correctly rounded answer to 2 or more
(θ =) 17(.4) (°) significant figure.
(Calculator displays: 17.35298907 for mg sin θ and
65g sin θ and 637.65 sin θ)
As an alternative to 17.35298907 may see Calculator
display or answer of:
• 17.34316751 for 638 sin θ =17(.3)
• 17.3515853 for 637.7 sin θ=17(.4)
22 • 17.28726034 for 640 sin θ =17(.3)
Common ecf:
(65gcos𝜃𝜃 = 190) = 72.6 (°) or 73(°) scores MP1 and MP2
(65 tan𝜃𝜃 = 190) = 71.1 (°) or 71(°) scores MP1 and MP2
Use of W = Fs cos𝜃𝜃 is only acceptable as an ecf where F
= 65g and component of weight is given as 65gcos𝜃𝜃 (or
equivalent) in 07.1
Alternative MP1:
height gain per second = 0.486 m and distance along the
slope per second = 1.63 m
OR
height gained per second
Use of sin θ =
distance travelled per second
07.3 3 3 x AO3
Less (useful) power output General marking principle:
Same gain in (gravitational) potential energy (in climbing hill) / same amount MP1 less (useful) power output
of work done (in climbing hill) / gains same height (in climbing hill)
MP2 basic point
Gains less (gravitational) potential energy every second
MP3 explains consequences of basic point in terms
OR (component of weight doing work against) of power (MP3 is an extension of MP2, quoting
Less (useful) power output ∆𝑊𝑊
P = without linking to an appropriate explanation is
∆𝑡𝑡
insufficient).
Effective θ has decreased / mg sin θ has decreased / component of the
weight parallel to the slope has decreased
Smaller force does less work per second Loses MP1: where conflicting statements made
about (useful) power output / states more power
OR (component of vertical velocity) output / total power output is same
Less (useful) power output
Loses MP3 for conflicting statements made in
support of explanation.
The vertical component of the velocity has decreased / height gained per
second decreases Accept θ as the effective angle to the slope.
∆𝑊𝑊
(P =) mg v sin θ has decreased / P = has decreased / less work done
∆𝑡𝑡
(against the weight) per second / Less gain in (gravitational) potential energy
per second 𝐸𝐸
Condone P= has decreased as MP3
𝑡𝑡
OR (distance travelled)
Less (useful) power output Treat ‘inputs more energy’ or ‘does more work’ as
Less force is exerted over greater distance (for same change in height) neutral
Smaller force does less work per second
07.4 Draws tangent which touches curve between 9 and 11 s Must see an attempt to draw a tangent to 3 1 x AO1
curve to score any marks.
1 x AO2
1 x AO3
Determine gradient of a tangent drawn at 5s / Determines
gradient of tangent drawn at 10s Read-offs must be within ½ square of accuracy
Condone one read-off error.
For tangent at t =5s, expect to see an answer
of 0.61 to 0.71 (m s−2). MAX 2 marks for this.
Accept answers in range 0.15 to 0.27 (m s−2)
Accept 2 or 3 significant figures only.
(acceleration =) 0.21 (m s−2)
MAX 1 mark
Condone a correctly determined gradient for a
tangent to the curve at any other point
between 5 and 11 seconds.
Condone ‘frictional forces increase with speed’ 25
07.5 Air resistance increases (with speed) / resistive forces 4 1 x AO1
Treat kinetic energy is transferred from the
increase (with speed) / Energy is transferred from the cyclist
cyclist as neutral. 3 x AO2
(due to work done) by resistive forces
MAX 3 from:
Initially, any of the gravitational potential energy that is
transferred is transferred to kinetic energy of cyclist
As speed increases, less of the gravitational potential energy
transferred per second is transferred to kinetic energy of
cyclist
As speed increases, energy transferred per second to the air
increases / as the speed increases, the energy transferred per The answer must be written in terms of energy
second from the cyclist increases transfers
At top speed, the gravitational potential energy that is
transferred (per second) is transferred to the air / the
gravitational potential energy (transferred per second) is being
transferred (from the cyclist) due to work done by resistive
forces
Total 13
How to answer it
Cyclist Dynamics, Power and Energy Transfers
What this question tests
This multi-part mechanics question evaluates your ability to resolve forces on an inclined plane, link mechanical power to velocity P = Fv , analyze vector components during angled paths, calculate acceleration from velocity-time graphs using tangents, and describe energy conservation/dissipation for terminal velocity.
Component of Weight Parallel to Slope
✅ Correct Answer
640 sin(θ) (or mg sin(θ) , 65g sin(θ) , 638 sin(θ) , 637.7 sin(θ) , 637.65 sin(θ) )
💡 Key Knowledge
- Weight always acts vertically downwards ( W = mg ).
- On a slope angled at θ to the horizontal, the component pulling the object down the slope is mg sin(θ) .
- Using m = 65 kg and g = 9.81 m s⁻² gives 637.65 N (or 638 N using g = 9.8 , or 640 N using g = 9.81 rounded to 2 s.f.).
❌ Common Errors
- Writing W sin(θ) without defining W as mg will lose the mark.
- Using cosine instead of sine ( mg cos(θ) is the component perpendicular to the slope).
Calculating Angle θ from Power and Speed
✅ Correct Answer
θ = 17.4° (Accept 2 or more significant figures, e.g., 17° )
📐 Step-by-Step Calculation
- Identify relevant equations: Power P = Fv and driving force up the hill equals the component of weight down the hill: F = mg sin(θ) .
- Substitute P = Fv : 310 = F × 1.63 → F = 190.18 N .
- Rearrange for sin(θ): sin(θ) = F / (mg) = 190.18 / (65 × 9.81) .
- Evaluate: sin(θ) = 0.2982 → θ = sin⁻¹(0.2982) = 17.4° .
🧠 Exam Technique & Error Traps
- Error Carry Forward (ECF): If you got part 07.1 wrong, use your expression here.
- Common Trap: Using cos(θ) or tan(θ) by mistake yields 72.6° or 190° (flagged in mark scheme as common errors). Ensure trigonometric functions match component resolutions.
Analyzing Power on a Zig-Zag Path
✅ Correct Answer
The useful power output is less than in Question 07.2.
💡 Key Physics Principles
- Taking a zig-zag path increases the distance traveled to gain the same vertical height, meaning the effective slope angle θ decreases.
- Since the effective angle decreases, the component of weight acting down the slope ( mg sin(θ) ) decreases.
- A smaller force is required to maintain the same steady speed, so less work is done per second ( P = Fv ), resulting in lower useful power output.
❌ Common Errors & Examiner Notes
- Simply stating "less power" without explaining *why* (linking to reduced effective angle, smaller component of weight, or less force needed) loses explanatory marks.
- Conflicting statements (e.g., stating power is less while claiming force is greater) forfeit mark point 1.
• MP1: States useful power output is less.
• MP2: Explains effective angle / component of weight parallel to slope / force required is smaller.
• MP3: Links P = Fv or W / t to show less work done per second.
Determining Acceleration from a Velocity-Time Graph
✅ Correct Answer
Acceleration = 0.21 m s⁻² (Acceptable range: 0.15 to 0.27 m s⁻² )
📐 Step-by-Step Calculation
- Locate target time: Find t = 10.0 s on the horizontal axis of Figure 10.
- Construct a tangent: Draw a straight line touching the curve precisely at t = 10 s with equal steepness above and below the point.
- Calculate gradient: Pick two points far apart on your tangent line:
Gradient = Δv / Δt = (v₂ - v₁) / (t₂ - t₁) . - Result: Expect an answer around 0.21 m s⁻² (tolerated range accounts for manual tangent variations).
🧠 Exam Technique
- Always use a sharp pencil and a transparent ruler for tangents.
- Choose points on the tangent line that span a large vertical and horizontal distance to minimize percentage reading errors.
• MP1: Draws a valid tangent touching the curve at t = 10 s (between 9s and 11s).
• MP2: Correctly determines the gradient of that tangent using a large triangle.
• MP3: Final value falls within 0.15 to 0.27 m s⁻² given to 2 or 3 s.f. with correct units ( m s⁻² ).
Energy Transfers and Terminal Velocity Graph Shape
✅ Correct Answer Summary
As speed increases, air resistance increases. Initially, lost gravitational potential energy ( GPE ) mostly transfers to kinetic energy ( KE ) causing acceleration. As speed increases, more energy per second transfers to thermal energy in the air via resistive forces, until energy transferred per second to air equals loss of GPE per second (terminal velocity).
💡 Key Marking Points (Max 3 from bulleted criteria + 1 for air resistance)
- Air resistance effect: Air resistance / resistive forces increase as speed increases. (Required point).
- Initial phase: Initially, almost all transferred GPE goes into increasing KE (steep initial gradient).
- Transition: As speed increases, a greater fraction of energy per second is transferred to the air (thermal/work done against drag) rather than increasing KE .
- Terminal state: At top speed (terminal velocity), 100% of the gravitational potential energy transferred per second goes into overcoming resistive forces (acceleration drops to zero, curve flattens).
❌ Common Errors
- Talking purely about forces (e.g. "weight equals air resistance") without explicitly discussing energy transfers and rates of transfer per second as demanded by the question.
Topics
Physics · 3.4 Mechanics and materials
Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.