AQA AS Level Physics Paper 2, June 2022: Question 4

10 marks · Medium difficulty · Short Answer

Calculate the acceleration, speed, impulse, distance travelled, and mass of an athlete in a chair sliding down an inclined rail and interacting with a force platform.

Practise this question

Question

Figure 6 shows an athlete strapped into a chair on a rail inclined at 30 degrees to the floor, facing a force platform inclined at 60 degrees. Figure 7 is a graph of force F against time t from 0.00 s to 2.25 s showing five key instances A, B, C, D, and E describing the athlete's motion and interaction with the platform.
Question text

04 Figure 6 shows apparatus used to measure the force exerted by an athlete during

a single-leg jump.

Figure 6

In Figure 6, the athlete is strapped into a chair and held at rest halfway along a rail.

The chair is then released to slide down the rail. The athlete keeps her right leg

extended until her right foot makes contact with a force platform.

Friction between the rail and the chair is negligible.

initial distance between right foot and platform = 0.30 m

angle between rail and floor = 30°

angle between platform and floor = 60°

04.1 Show that the athlete and chair accelerate towards the platform at

approximately 5 m s−2.

[1 mark]

04.2 Calculate the speed of the athlete when her right foot makes initial contact with the

platform.

[2 marks]

18 −1

speed = m s

After her right foot makes contact with the platform, she uses her right leg to stop

moving and then push herself back up the rail. She slides down the rail again, lands

on the platform with both feet and comes to rest.

Figure 7 shows the variation of force F on the platform with time t during the full

motion.

Figure 7

The sequence below describes what happens at the five instances A, B, C, D and E

shown in Figure 7.

A: athlete and chair are released at t = 0.00 s

B: right foot of athlete contacts the platform with leg fully extended

C: right foot loses contact with the platform

D: athlete lands on the platform with both feet

E: athlete and chair come to rest

04.3 Determine the impulse provided by the force platform between B and C.

[2 marks]

19 impulse = N s

04.4 Determine the distance travelled by the athlete between C and D.

[3 marks]

distance travelled = m

04.5 Determine, using Figure 7, the combined mass of the athlete and chair.

[2 marks]

mass = kg

END OF SECTION B

Section C

Each of Questions 05 to 34 is followed by four responses, A, B, C and D.

For each question select the best response.

Only one answer per question is allowed.

*For each question, completely fill in the circle alongside the appropriate answer.19*

CORRECT METHOD WRONG METHODS

If you want to change your answer you must cross out your original answer as shown.

If you wish to return to an answer previously crossed out, ring the answer you now wish to select

as shown.

You may do your working in the blank space around each question but this will not be marked.

Do not use additional sheets for this working.

Mark scheme

Show the mark scheme Mark scheme providing answers and guidance for sub-questions 0.4.1 through 0.4.5, detailing required equations, area calculations from graphs, and accepted numerical ranges.

Question Answers Additional Comments/Guidance Mark AO

04.1 a = 9.81 sin30 = 4.9 m s–2 ✓ Allow g sin30 or 9.8sin30 1 AO2

( ) ( ) seen

Accept cos60

04.2 substitutes into v2 = u2 + 2as e.g. v2 = 2 × 5 × 0.3 Do not allow 9.81 for a in suvat equation. 2 2 ×

2 AO2

OR uses v = g 0.3 cos 60 ✓

1.7 (m s−1) ✓

– SICS – – JUNE 2022

04.3 attempt to find area between 0.35 s (B) and 0.80 s (C) 1✓ Max 1 for counting (small) squares AND a 2 1 x AO3

conversion factor of 2 N s stated

1 x AO2

Do not allow use of approximated shapes.

For 1✓ need to see working for at least one part of

the area under the graph.

May see:

Triangle: 0.5×0.05×1100 = 27.5

Trapezium: 0.5×(1100+1300)×0.10 = 120

Trapezium: 0.5×(1000+1300)×0.15 = 172.5

Triangle: 0.5×0.15×1000 = 75

Treat “400” as a 2 sf answer.

answer in range 390 to 400 (N s) 2✓

– SICS – – JUNE 2022

04.4 uses a relevant time in suvat equation(s) to get s 1✓ For 1✓ condone 9.81 for a. 3 2 × AO3

Do not allow 1✓ or 3✓ for solutions that use

-1 1 x AO2

u=1.7 m s

For 1✓ allow 0.60 to 0.63 s for duration C to D.

1. Direct method: 𝑠 = 2 𝑎𝑡

2. a) Obtains u first using 𝑣 = 𝑢 + 𝑎𝑡 OR 17

𝑠 = 𝑢𝑡 + 𝑎𝑡

2. b) Then s using 𝑠 = (𝑢 + 𝑣)𝑡 OR

𝑣2 = 𝑢2 + 2𝑎𝑠

doubles their s OR halves their C to D duration 2✓ Expect to see u = 1.5 (m s–1)

answer that rounds to 0.5 (m) 3✓ For 3✓ accept 0.44 (m).

04.5 reads resting force from graph = 360 N OR divides an for ✓ allow use of their 04.3 with v = 3.2 m s-1 2 1 x AO3

incorrect reading by 5 (4.91 N/kg) 1✓ 1 x AO2

72 or 73 (kg) 2✓

Total 10

How to answer it

Mechanics of an Athlete Jumping on a Force Platform

AQA AS Level Physics • Section B Exam Practice

What this question tests

This multi-part question tests core mechanics topics including resolution of forces on inclined planes, constant acceleration equations (SUVAT), impulse calculation as the area under a force-time graph, kinematics of unpowered descent, and interpreting graphical data to determine dynamic and static weight forces ($W = mg$).

Question 04.1 • Acceleration on an Incline

Show that the athlete and chair accelerate towards the platform at approximately 5 m s⁻².

💡 Key Knowledge

  • On an incline at angle θ to the horizontal, the component of gravitational acceleration acting down the slope is given by g sin(θ) .
  • Friction is stated as negligible, meaning only gravity drives the motion down the rail.

📐 Calculation Steps

  1. Identify standard gravitational field strength g = 9.81 m s⁻² (or 9.8 m s⁻²).
  2. Identify the inclination angle θ = 30° .
  3. Calculate component: a = 9.81 × sin(30°) = 4.905 m s⁻² .
Mark Scheme Guidance: 1 mark awarded for seeing 9.81 sin(30) = 4.9 (m s⁻²) (or equivalent using 9.8 or cos 60°).

Question 04.2 • Velocity Before Impact

Calculate the speed of the athlete when her right foot makes initial contact with the platform.

🧠 Exam Technique

Since acceleration is constant, you can apply standard SUVAT equations or equate gravitational potential energy to kinetic energy.

📐 Calculation Steps

  1. Note down known variables: u = 0 , a = 5 m s⁻² (or 4.9 m s⁻²), s = 0.30 m .
  2. Select SUVAT equation: v² = u² + 2as .
  3. Substitute values: v² = 2 × 5 × 0.30 = 3.0 .
  4. Square root: v = √3.0 = 1.732... m s⁻² .

❌ Common Errors

  • Using free-fall acceleration g = 9.81 m s⁻² directly for a without resolving it down the 30° slope will cost you marks.
✅ Correct Answer: 1.7 m s⁻¹ (2 sig fig standard format).

Question 04.3 • Impulse from a Graph

Determine the impulse provided by the force platform between B and C.

💡 Key Knowledge

Impulse is defined as the change in momentum, which is equal to the area under a force-time graph.

📐 Step-by-Step Analysis

  1. Locate time coordinates on Figure 7: point B is at t = 0.35 s and point C is at t = 0.80 s .
  2. Count squares or split the region under the curve between B and C into geometric shapes (triangles and trapeziums).
  3. Account for axis scaling carefully: each small square has an area value determined by its width and height units ( 2 N s per large square grid block).
  4. Sum the calculated areas to land in the acceptable range.

❌ Common Errors

Do not use crude approximations like a single large triangle; examiners require careful breakdown into sub-shapes or accurate square counting.

✅ Correct Answer: Any value in the range 390 to 400 N s .

Question 04.4 • Kinematics After Contact

Determine the distance travelled by the athlete between C and D.

🧠 Exam Technique

Between C and D, the force from the platform drops to zero. The athlete is in unpowered projectile/coasting motion up the rail under gravity alone.

📐 Step-by-Step Calculation

  1. Identify initial velocity at point C: this equals the final velocity from the deceleration phase, which is u = 1.5 m s⁻¹ (or using 1.7 m s⁻¹ from earlier parts safely via carried-through error).
  2. Determine time interval between C and D from Figure 7: duration is approximately 0.60 to 0.63 s .
  3. Apply SUVAT: s = ut + (1/2)at² where deceleration a = -4.9 m s⁻² , OR use average velocity s = (1/2)(u + v)t .
✅ Correct Answer: 0.5 m (accept 0.44 m depending on precise time coordinate reading).

Question 04.5 • Mass Determination

Determine, using Figure 7, the combined mass of the athlete and chair.

💡 Key Knowledge

When the system is at rest on the platform (before release at A or after coming to rest at E), the force reading on the platform equals the total weight W = mg .

📐 Calculation Steps

  1. Read the resting force F from Figure 7 where the graph flattens out at the end (point E): F = 360 N .
  2. Rearrange weight formula for mass: m = W / g .
  3. Calculate: m = 360 / 4.91 (or divided by 9.81 / 5.0 depending on context track).
  4. Result: 360 / 5.0 = 72 kg (or 73 kg ).

❌ Common Errors

Students often mistake peak impact forces for the static weight value. Always look at the steady tail-end of the graph (point E) where motion has ceased.

✅ Correct Answer: 72 kg or 73 kg .

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.