AQA AS Level Physics Paper 1, June 2025: Question 5

15 marks · Hard difficulty · Short Answer

Calculate the cable extension using the Young modulus and resolving forces, determine gravitational potential energy and cable tension from a velocity–time graph, find the impact velocity of a detached object, and sketch its kinetic energy–time graph.

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Question text

05 Figure 6 shows a stationary lift suspended using three steel cables P, Q and R.

The centre of mass of the lift is vertically below cable P.

Figure 6

The mass of the lift is 850 kg.

The angle between Q and R is 75°

Q and R each have an unstretched length of 2.0 m.

The cross-sectional area of each cable is 1.4 × 10−4 m2.

The Young modulus of the steel is 2.1 × 1011 Pa.

05.1 Determine the extension of Q.

[4 marks]

extension = m

Figure 7 is a velocity–time graph for the lift as it travels upwards from ground level.

Figure 7

05.2 Determine the gravitational potential energy gained by the lift during the first 1.5 s of

the motion.

[3 marks]

gravitational potential energy17 = J

05.3 Q and R are connected to P as shown in Figure 6.

*16*Determine the tension in P during the last 2.0 s of the motion.

[3 marks]

tension = N

After 4.0 s of the motion, the lift is moving at a constant velocity and has travelled

5.5 m from ground level. An object becomes detached from the underside of the lift.

05.4 Determine the velocity of the object when it reaches ground level.

[3 marks]

velocity = m s−1

05.5 Sketch on Figure 8 the variation with time of the kinetic energy of the object from

when it becomes detached until it hits the ground.

No calculations are required.

*17* [2 marks]

Figure 8

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidance Mark AO

05.1 • Calculation of the weight MP1 e.g. 4 2 ×

AO1

(W =) 850g or 8340 (N) or 850 × 9.81 seen or

2 ×

calculating the weight of 425 kg.

AO2

MP2 is for applying equilbrium condition

• 2 T cos 37.5 = their W using their weight.

OR Condone one other error, must see an

attempt to resolve that includes the tension.

2T sin 52.5 = their W

i.e. using W sin 𝜃𝜃 or W cos 𝜃𝜃 does not get

this mark

𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑓𝑓𝑓𝑓𝑒𝑒𝑓𝑓𝑒𝑒 ×2.0

• use of Young modulus = 1.4 ×10−4 ×𝑒𝑒𝑒𝑒𝑡𝑡𝑒𝑒𝑛𝑛𝑒𝑒𝑒𝑒𝑓𝑓𝑛𝑛 MP3 for use of Young modulus equation.

Condone power of ten errors

their 𝐹𝐹 their σ ∆L

Use of 𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝜎𝜎 = AND E= AND their ε= Their stress and their strain must be

𝐴𝐴 ε L

consistent with their force.

Condone power of ten errors

3.6 × 10-4 (m) Using weight as tension: ∆L =5.7 x10-4(m)

obtains 2 marks

Accept any answer that rounds to

3.6 × 10−4 (m)

05.2 3 2 × AO1

Expect to see ∆h = 1.275 (m)

determine correct area under graph for correct region with Allowable range for v: 1.625 to 1.775 (ms-1) 1 × AO2

their read-off for v

Allowable range for ∆h : 1.21875 to 1.33125 (m)

OR determines ∆h using appropriate suvat equation

Allow their ∆h in MP2

Use of ∆Ep = mg∆h

Condone answer in range 1.04958 × 104 (J) to

1.0965 × 104 (J) rounded to 2 or more significant

1.1 × 104 (J)

figures.

05.3 Any two from 3 AO2

• acceleration from the graph = (−) 0.85 (m s−2)

• uses F = ma with their acceleration

• subtracts their (resultant) force from their weight Check their weight against value seen in

working in 05.1 (treat as ecf)

22 OR

850g −𝑇𝑇 = 850 × their acceleration or equivalent seen 3rd bullet point, 2nd option is worth 2nd and 3rd

bullet points

7600 (N)

05.4 uses initial velocity = 1.7 (m s−1 upwards) 3 AO2

Use of v2 = u2 + 2as with their u Expect to see v2 = 1.72 + 2 x 9.81 x 5.5

MP2 Condone incorrect signs seen in

attempted use of v2 = u2 + 2as

Do not penalise inconsistent signs in MP2 if

Use of equivalent combination of suvat equations combination of suvat equations used

with their u

Calculator value 10.52615789

−1 OR

11 (m s )

Calculator value 10.54943126 (for student using

s = 5.525 m using area under v−t graph in

Figure 7)

05.5 • correct general shape with minimum turning point of zero 2 AO3

(displaced to left) (1)

• lower initial non-zero KE than final KE with a minimum

turning point inbetween (where final KE is KE just before

impact with ground) (2)

Need to see 5.4 for possible ecf

Max 1 for ECF from 5.4 where u is indicated as being zero

• correct general shape beginning at (0,0) with increasing

gradient for KE with respect to time (3)

Max 1 ECF from 5.4 where u is indicated as being

downwards and non-zero

• correct general shape beginning at positive y-intercept with

increasing gradient for KE with respect to time (4)

Total 15

How to answer it

Mechanics, Materials & Vertical Motion of a Cable-Suspended Lift

📌 What this question tests

This multi-topic question assesses your mastery of core AS-Level mechanics and properties of materials:

Statics & Resolving Forces: Equilibrium of non-parallel coplanar vectors using trigonometry.
Young Modulus & Materials: Stress, strain, and cable extension using E = (F · L) / (A · ΔL) .
Kinematics & Graphical Analysis: Determining distance from area under a v–t graph and calculating Gravitational Potential Energy ( ΔE_p = mgΔh ).
Newton’s Second Law in Non-Equilibrium Systems: Resultant forces during deceleration ( W - T = ma ).
SUVAT & Free-Fall Energy Analysis: Velocity of an object released with non-zero upward velocity and sketch of parabolic E_k vs t curves.

Part 05.1: Cable Tension & Extension (Young Modulus)

4 Marks | AO1, AO2

📐 Step-by-Step Calculation

Find total weight of lift:
W = mg = 850 × 9.81 = 8338.5 N (or 8340 N using 9.81 ).
Resolve vertically for equilibrium:
The angle between Q and R is 75°, so the angle each cable makes with the vertical is θ = 75° / 2 = 37.5° .
2 · T · cos(37.5°) = W
T = 8338.5 / (2 × cos(37.5°)) = 5255 N
Apply Young Modulus equation:
E = (T · L) / (A · ΔL) ⇒ ΔL = (T · L) / (A · E)
ΔL = (5255 × 2.0) / (1.4 × 10⁻⁴ × 2.1 × 10¹¹)
Calculate extension:
ΔL = 10510 / (2.94 × 10⁷) = 3.57 × 10⁻⁴ m ≈ 3.6 × 10⁻⁴ m

❌ Common Errors

Using full angle: Using cos(75°) instead of half-angle cos(37.5°) .
Using total weight directly as tension: Forgetting that tension is shared by two angled cables. Setting T = W gives ΔL = 5.7 × 10⁻⁴ m (loses 2 marks).
Missing the factor of 2: Writing T cos(37.5°) = W rather than 2T cos(37.5°) = W .

Mark Scheme Breakdown:
• MP1: Correct calculation of weight ( 850 × 9.81 = 8340 N or 850g ).
• MP2: Correct equilibrium condition: 2T cos(37.5°) = W or 2T sin(52.5°) = W .
• MP3: Correct substitution into Young Modulus equation: ΔL = (T × 2.0) / (1.4 × 10⁻⁴ × 2.1 × 10¹¹) .
• MP4: Final answer of 3.6 × 10⁻⁴ m (accepts any rounding to 3.6 × 10⁻⁴ ).

Part 05.2: Gravitational Potential Energy Gain

3 Marks | AO1, AO2

📐 Step-by-Step Calculation

Find vertical height gained (Δh):
Area under v–t graph from t = 0 to t = 1.5 s (triangle):
v = 1.7 m s⁻¹ at t = 1.5 s
Δh = ½ × base × height = 0.5 × 1.5 × 1.7 = 1.275 m
Calculate potential energy gained:
ΔE_p = mgΔh
ΔE_p = 850 × 9.81 × 1.275
Evaluate:
ΔE_p = 10 631 J = 1.1 × 10⁴ J (to 2 s.f.)

🧠 Exam Technique

Reading grids accurately: Check the axis minor grid divisions. On this axis, 5 small squares = 0.2 m s⁻¹, so 1 small square = 0.04 m s⁻¹. At t = 1.5 s , v = 1.70 m s⁻¹ .
Allowable ranges: The mark scheme allows v from 1.625 to 1.775 m s⁻¹, giving Δh between 1.22 m and 1.33 m, and final energy within 1.05 × 10⁴ J to 1.10 × 10⁴ J .

Mark Scheme Breakdown:
• MP1: Correct area under graph or valid SUVAT calculation ( Δh = 1.275 m ).
• MP2: Correct use of ΔE_p = mgΔh using student's Δh .
• MP3: Final answer: 1.1 × 10⁴ J (condone 1.05 × 10⁴ to 1.10 × 10⁴ J to 2 or more s.f.).

Part 05.3: Cable Tension During Deceleration

3 Marks | AO2

📐 Step-by-Step Calculation

Find acceleration during the last 2.0 s:
Deceleration occurs from t = 7.5 s to t = 9.5 s ( Δt = 2.0 s ).
a = (v - u) / t = (0 - 1.7) / 2.0 = -0.85 m s⁻²
Formulate Newton's Second Law:
Lift is moving upwards but decelerating downwards (resultant force acts downwards):
W - T = ma ⇒ T = W - ma = m(g - a)
Alternatively: T - W = m(-0.85) ⇒ T = 8338.5 - (850 × 0.85)
Calculate tension in cable P:
T = 8338.5 - 722.5 = 7616 N ≈ 7600 N

❌ Common Errors

Wrong sign on acceleration: Adding ma to weight instead of subtracting it. Because the lift is decelerating while rising, cable tension must be less than the lift's stationary weight.
Dividing by 2 or resolving: Cable P supports the entire assembly vertically, so no trigonometric resolution is needed for P!

Mark Scheme Breakdown: (Any two from first three points, plus final answer)
• Acceleration from graph = (-) 0.85 m s⁻² .
• Uses F = ma with their acceleration.
• Subtracts resultant force from weight: 850g - T = 850 × a .
• Final tension: 7600 N .

Part 05.4: Velocity of Detached Object Hitting Ground

3 Marks | AO2

📐 Step-by-Step Calculation

Identify initial conditions:
At t = 4.0 s , the lift is at constant cruise speed.
Initial velocity of object: u = +1.7 m s⁻¹ (upwards).
Displacement to ground: s = -5.5 m (downwards).
Acceleration: a = -9.81 m s⁻² (downwards).
Apply SUVAT equation:
v² = u² + 2as
v² = (1.7)² + 2 × (-9.81) × (-5.5)
v² = 2.89 + 107.91 = 110.8
Solve for speed/velocity:
v = √(110.8) = 10.53 m s⁻¹ ≈ 11 m s⁻¹

🧠 Exam Technique

The "moving platform" trap: When an object detaches from a moving lift, its initial velocity is not zero—it shares the velocity of the lift ( 1.7 m s⁻¹ upwards ).
Alternative Conservation of Energy method:
½mv² = ½mu² + mgh ⇒ v = √(u² + 2gh) . Notice mass cancels out immediately!

Mark Scheme Breakdown:
• MP1: Initial velocity identified as 1.7 m s⁻¹ upwards.
• MP2: Correct use of v² = u² + 2as with their u (condoning sign errors in substitution).
• MP3: Final answer: 11 m s⁻¹ (accepts 10.5 m s⁻¹).

Part 05.5: Kinetic Energy vs Time Sketch

2 Marks | AO3

💡 Physical Analysis of the Motion

Initial instant ( t = 0 ): The object starts with an upward velocity of 1.7 m s⁻¹ , so initial E_k > 0 (positive y-intercept).
Rising to peak: As it rises under gravity, velocity decreases to zero at the highest point. Thus, E_k drops to exactly zero at a turning point.
Falling to ground: It falls back down, accelerating under gravity. Because it falls below its release point ( s = -5.5 m ), its impact speed ( 10.5 m s⁻¹ ) is much greater than its initial speed ( 1.7 m s⁻¹ ). Therefore, final E_k >> initial E_k .
Curve shape: Since v = u - gt , kinetic energy is quadratic in time: E_k = ½m(u - gt)² , which is an upward-opening parabola touching the time axis.

✅ Required Sketch Description

Y-intercept: Starts above the origin on the kinetic energy axis at a moderate non-zero value.
Minimum: Curves downwards smoothly to touch the time axis ( E_k = 0 ) displaced to the left of the graph.
Final point: Curves upward with an increasing gradient, reaching a final E_k value significantly higher than the initial starting point.

Mark Scheme Breakdown:
• Mark 1: Parabolic curve with minimum turning point touching zero ( E_k = 0 ), displaced to the left.
• Mark 2: Initial non-zero E_k strictly lower than the final E_k at impact.
Note: If student assumed u = 0 in 05.4, ECF allows max 1 mark for a curve starting at (0,0) with increasing gradient.

Topics

Physics · Practical skills · 3.4 Mechanics and materials · Data analysis

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.