AQA A-Level Mathematics Paper 2, June 2025: Question 16
12 marks · Medium difficulty · Multi-step Problem
Calculate the tension and acceleration of a sledge pulled up an inclined plane at an angle to the slope, and state an assumption used in both models.
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Question text
16 In this question use g = 9.8 m s–2
A sledge is pulled in a straight line up a rough path, as shown in the diagram.
The path is inclined at an angle of 20° to the horizontal.
The sledge is pulled by a light, inextensible rope inclined at an angle of 30° to the path.
30°
20°
The mass of the sledge is 10 kilograms.
16 (a) In one model, the sledge moves at a constant speed and experiences a combined
resistance force of 15 newtons.
Find the tension in the rope for this model.
[4 marks]
16 (b) In a different model, the sledge experiences no air resistance.
The tension in the rope for this model is 54 N
(24)
The coefficient of friction between the sledge and the path is 0.2
Find the acceleration of the sledge for this model.
[7 marks]
16 (c) State an assumption you have used to answer both parts (a) and (b).
[1 mark]
Mark scheme
Show the mark scheme
Q Marking instructions AO Marks Typical solution
16(a) Obtains 10gsin20 OE 3.1b B1 10gsin20 +15 = Tcos30
Resolves parallel to the path to
form a three-term equation with 3.3 M1
at least two correct terms T = 56 02.
Obtains T = 56 N
OE 1.1b A1
10gsin20 +15 = Tcos30
Obtains AWRT 56
1.1b A1
Condone missing units.
Subtotal 4
16(b) Uses F = R 1.1b B1 R + 54sin 30 = 10g cos20
Obtains 10g cos20 OE 1.1b B1 R = 65 0899. N
Resolves perpendicular to the 3.3 M1
plane to form a three-term
dimensionally correct equation 54cos30 − R −10g sin 20 = 10a
with at least two terms correct.
Obtains a correct expression or 1.1b A1 54cos30 − 0 2. 65 0899. − 98sin 20
value for R AWRT 65 a =
PI by friction = AWRT 13 02. 10
Uses F = ma to form a four-term 3.4 M1
dimensionally correct equation, a = 0 023. ms−2
excluding a 15N force, with at
least three terms correct.
FT their R value
Condone inclusion of a 15N
force from 16(a) as a 5th term
Forms fully correct equation in a 1.1b A1F
for the particle with all values
substituted.
FT their R value
Note this alone obtains
B1B1M1A1M1A1
Obtains AWRT 0.023 m s–2 1.1b A1
Subtotal 7
16(c) States the sledge is modelled as 3.5a E1 The sledge is modelled as a particle.
a particle
If more than one assumption is
stated E0
Subtotal 1
Question 16 Total 12
How to answer it
Forces on an Inclined Plane & Newton's Laws
This question assesses your ability to set up and solve mechanics problems involving inclined planes and angled tension forces:
- Resolving forces at angles: Splitting non-axial forces into perpendicular and parallel components along an inclined plane.
- Equilibrium condition: Interpreting "constant speed" as zero acceleration ( a = 0 ) and applying ΣF = 0 parallel to the slope.
- Normal reaction with angled pull: Recognising that tension acts partly away from the surface, reducing normal contact force R .
- Limiting friction & dynamics: Applying F = μR and Newton's Second Law ( F = ma ) up the plane.
- Modelling assumptions: Stating the precise physical implications of modelling an extended body as a particle.
- Weight ( W = mg = 98 N ): Acts vertically downward. Components relative to the 20° slope: 98 cos 20° perpendicular into slope, 98 sin 20° down the slope.
- Tension ( T ): Pulls at 30° above the incline. Components: T cos 30° up the slope, T sin 30° perpendicular away from the slope.
- Normal Reaction ( R ): Acts perpendicular to the slope, pointing outward from the surface.
- Resistance / Friction: Opposes the upward motion, acting directly down along the plane.
Question 16 (a)
Model 1: Constant speed with combined resistance force [4 Marks]
📐 Step-by-Step Calculation
- Identify parallel equilibrium: Constant speed implies acceleration a = 0 . Therefore, total force up the slope equals total force down the slope.
- Component of weight down the plane:
Wparallel = mg sin 20° = 10(9.8) sin 20° = 98 sin 20° N - Balance parallel forces:
T cos 30° = 10g sin 20° + 15
T cos 30° = 33.518 + 15 = 48.518 - Solve for T:
T = 48.518 / cos 30° = 48.518 / 0.8660 = 56.02... N
✅ Correct Answer
T = 56 N (or 56.0 N)
• [B1] Obtains 10g sin 20° (or 98 sin 20° )
• [M1] Resolves parallel to the path into a 3-term equation with at least 2 correct terms
• [A1] Obtains fully correct equation: 10g sin 20° + 15 = T cos 30°
• [A1] Obtains AWRT 56 (units optional)
🧠 Exam Technique
- Notice that in part (a), you do not need to resolve perpendicular to the slope! The resistance is given directly as a constant 15 N , not as μR . Resolving perpendicular would waste valuable time.
- Always check your calculator is in degree mode before resolving trigonometry terms.
❌ Common Errors
- Angle confusion: Mixing up sin and cos for slope components. Weight component parallel to an inclined plane is always mg sin θ .
- Angle reference: Using 50° (20° + 30°) instead of keeping the components referenced directly to the slope coordinate axes.
Question 16 (b)
Model 2: Accelerating sledge with friction (μ = 0.2, T = 54 N) [7 Marks]
📐 Step-by-Step Calculation
- Resolve perpendicular to the slope (perpendicular equilibrium):
R + T sin 30° = mg cos 20°
R + 54 sin 30° = 10(9.8) cos 20°
R + 27 = 92.0899...
R = 65.0899... N - Find maximum friction force ( F = μR ):
F = 0.2 × 65.0899... = 13.018... N - Apply Newton's Second Law up the slope ( ΣF = ma ):
T cos 30° - mg sin 20° - F = ma
54 cos 30° - 98 sin 20° - 13.018 = 10a - Evaluate numerical values:
46.765 - 33.518 - 13.018 = 10a
0.229... = 10a
a = 0.0229... m s⁻²
✅ Correct Answer
a = 0.023 m s⁻²
• [B1] Uses F = μR
• [B1] Obtains 10g cos 20°
• [M1] Resolves perpendicular to plane to form 3-term dimensionally correct equation (at least 2 correct terms)
• [A1] Correct value or expression for R (AWRT 65 N or friction AWRT 13.02 N)
• [M1] Uses F = ma parallel to plane (excluding 15 N) with at least 3 correct terms; FT their R
• [A1F] Fully correct equation in a with values substituted
• [A1] Obtains AWRT 0.023 m s⁻²
🧠 Exam Technique: Two-Stage Process
- Stage 1 (Perpendicular): Always resolve perpendicular to find R first. Crucially, the tension pulls upwards and forwards, so it lightens the normal reaction: R = mg cos 20° - T sin 30° .
- Stage 2 (Parallel): Use F = μR to find the opposing friction, then form the equation of motion along the plane: T cos 30° - mg sin 20° - F = ma .
❌ Critical Traps & Common Errors
- Forgetting tension component in R: Assuming R = mg cos 20° without subtracting 54 sin 30° . This is the single most common mistake on this topic!
- Carrying over the 15 N force: The question states this is a different model with no air resistance. The 15 N resistance from part (a) must not be included.
- Premature rounding: Rounding R to 65 N too early. Keep the unrounded value stored in your calculator to avoid rounding error in the final answer ( 0.023 ).
Question 16 (c)
Modelling Assumption [1 Mark]
✅ Correct Answer
"The sledge is modelled as a particle."
• [E1] States the sledge is modelled as a particle.
• Examiner Rule: If more than one assumption is stated, award 0 marks (E0).
💡 Why is this assumption valid for both parts?
Modelling the sledge as a particle means:
- All forces act at a single point, which prevents the sledge from rotating or toppling.
- Its mass is concentrated at a single point, ignoring the dimensions and shape of the sledge.
❌ Deadly Exam Trap: The "Shopping List" Penalty
The mark scheme explicitly specifies: "If more than one assumption is stated, E0."
Never write: "The sledge is a particle and g is constant" or "The sledge is a particle and rope is light/inextensible". Mentioning more than one assumption automatically invalidates your mark, even if both are true! Give exactly one clear assumption.
Topics
Mechanics · R: Forces and Newton’s laws
Question and mark scheme from the AQA A-Level Mathematics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.