AQA AS Level Physics Paper 1, June 2024: Question 4
17 marks · Medium difficulty · Short Answer
Calculate distances, velocities, forces, moments, and explain air resistance acting on a car and trailer system using equations of motion, Newton's laws, and the principle of moments.
Practise this questionQuestion
Question text
04 In this question, assume that all forces are coplanar.
04.1 Figure 7 shows a car and trailer moving at 25 m s−1 at a distance s from traffic lights.
Figure 7
The driver applies the brakes so that the car and trailer stop at the lights.
The car and trailer undergo a constant deceleration of 2.8 m s−2.
Calculate s.
[2 marks]
s = m
The car and trailer accelerate from rest along a horizontal road.
Figure 8 shows:
• that the car exerts a horizontal force PH on the trailer
• that the trailer has a weight of 2500 N
• the reaction force R on the wheels of the trailer.
Figure 8
04.2 Initially, there are no resistive forces and the trailer accelerates at 1.5 m s−2.
*12* Calculate the initial value of PH.
[2 marks]
14 P =
H N
Air resistance D acts on the trailer when it is moving. D increases as the velocity v of
the trailer increases.
Figure 9 shows how D varies with v.
Figure 9
Figure 10 shows the forces acting on the trailer when it is travelling at a constant
horizontal velocity.
The car exerts a vertical force PV and a horizontal force PH on the trailer when it is
travelling at a constant horizontal velocity v1.
An enlarged view of PV and PH is also shown in Figure 10.
Figure 10
The horizontal force P is now greater than the value calculated in Question15 04.2.
H
04.3 The vertical force PV is 762 N.
The resultant of PH and PV is 912 N.
Determine v1.
[3 marks]
v = m s−1
D can be considered to act at the position shown in Figure 11. For some of the
forces, the distances of their lines of action from the centre of the trailer’s wheel have
been included.
Figure 11
04.4 Explain why PH has no moment about the centre of the trailer’s wheel in Figure 11.
[1 mark]
04.5 When the car and trailer travel with velocity v2, PV is zero.
Determine v2.
[3 marks]
v = m s−1
04.6 The air resistance D acting on the trailer increases as the velocity v of the trailer
increases.
Explain this increase in D with reference to the momentum of the air displaced by
the trailer.
You should also refer to appropriate Newton’s laws of motion.
[3 marks]
04.7 The car has a maximum power output of 95 kW.
The maximum velocity of the car and trailer is 25 m s−1.
At this velocity, the force D on the trailer is 3100 N.
The car exerts a horizontal force PH on the trailer and the trailer exerts an equal and
opposite force of magnitude PH on the car.
Assume that air resistance and PH are the only resistive forces acting on the car.
Calculate the air resistance acting on the car when it is travelling at a constant velocity
of 25 m s−1.
[3 marks]
air resistance on car = N
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark AO
04.1 Use of appropriate equation(s) of motion For example: 2 1 × AO1
expect to see use of: 1 × AO2
• v2 = u2 + 2as
use of is:
rearrangement to make s subject.
𝑢𝑢2 𝑣𝑣2 − 𝑢𝑢2
𝑠𝑠 = 2𝑎𝑎 or 𝑠𝑠 = 2𝑎𝑎
OR substitution
condone one error in substitution.
𝑢𝑢+𝑣𝑣
• v = u +at and s = 2 𝑡𝑡
condone one error in substitution.
(s =) 112 m
Calculator display = 111.60714285̇
110 (m) correct to 2 sf
04.2 Use of W = mg to determine mass m = 254.8 (kg) 2 1 × AO1
OR 1 × AO2
Use of F = ma with their mass: allow use of m=2500 in F=ma
allow use of g = 9.8 N kg-1 (2 sf)
(PH =) 380 (N)
g = 9.8
calculator display: 382.653061224489
g = 9.81
calculator display: 382.2629969
04.3 Attempts to use Pythagoras’s theorem. Condone one error in attempt to use either 3 2 × AO2
Pythagoras’s theorem or trigonometric ratios: 1 × AO3
OR Substitution or rearrangement PH would be
subject.
PV
Attempts to use sin θ = and PH = PRes cosθ
PRes
PH = 500 N
Allow PH = 90√31
OR
A correct read-off of their v from Figure 9 for their PH Read-off within ± 𝑠𝑠𝑚𝑚𝑠𝑠𝑙𝑙𝑙𝑙𝑒𝑒𝑠𝑠𝑡𝑡 𝑑𝑑𝑒𝑒𝑑𝑑𝑒𝑒𝑠𝑠𝑒𝑒𝑑𝑑𝑑𝑑 of their
accurate read-off.
Must see working to support answer from MP1
or MP2 to score all 3 marks.
Answer in range without MP1 or MP2 obtains
MP3 only.
Answer in range 9.75 to 10.25 (ms-1)
−1 Condone 10.3 (ms-1) to 3 sf
(v1 =) 10 (m s )
04.4 PH’s line of action passes through the centre of the wheel. Condone: 1 AO1.1b
OR Idea that PH acts through the centre of the
the perpendicular distance between P ’s line of action and trailer’s wheel.
H
the centre of the wheel is zero. Or
The perpendicular distance is zero.
OR
Do not accept:
m = Fd and d is zero
The distance between the centre of the wheel
and PH is zero.
PH acts parallel to the centre of the wheel is
insufficient.
04.5 Max 2 from: Condone one error in attempt to use: 3 1 × AO1
• Attempts to use principle of moments about the centre D × 0.95 = 2500 × 0.8 1 × AO2
of the trailer’s wheel.
1 × AO3
• D = 2100 (N) 1
Read-off within ± 𝑠𝑠𝑚𝑚𝑠𝑠𝑙𝑙𝑙𝑙𝑒𝑒𝑠𝑠𝑡𝑡 𝑑𝑑𝑒𝑒𝑑𝑑𝑒𝑒𝑠𝑠𝑒𝑒𝑑𝑑𝑑𝑑 of their
accurate read-off. Allow a read-off for a force
including PH as D may equal PH
• Read-off from graph of v2 for their value of D.
v2 must be greater than zero.
Must see working that includes a correct
principle of moments equation to score MP3.
MP3:
(Read-off from graph, v =) 20.5 (m s−1)
2 Answer in range 20 to 21 m s-1
04.6 As v increases: Compensatory mark, Max 1: 3 AO2
there is a greater force on the air (as v
more air particles are given momentum (each second). increases)
OR
each air particle given more momentum.
OR
Idea that more air is displaced (each second)
OR
Idea that the displaced air has a greater velocity. 21
OR
More air particles change direction (per second)
OR
There are more collisions with the air (particles each second)
Idea of a greater rate of change of momentum of air requires
a greater force on air (relates to Newton’s 2nd law) To achieve 3 marks, must link:
MP2 to Newton’s 2nd law or its formula
OR
(Greater) force on air by trailer means (greater) force on trailer
by air (relates to Newton’s 3rd law.) MP3 to Newton’s 3rd law
04.7 Max 2 from Condone one error in use of P=Fv 3 1 × AO1
P = Fv Where: 2 × AO2
• Use of
P and v
or
F and v
22 have been substituted.
Expect to see:
• 95 × 103 = F × 25
• (F =) 3800 (N)
• (P =) 3100 × 25
• (3100 × 25=) 77500
• Subtracts D from their thrust.
3800 − D OR 3800 − 3100
• Subtracts their rate of work done by D on trailer from
95 kW. 3
95 × 10 – 77500
OR
17500
OR
95 × 103 – their rate of work done by D
(Air resistance on car =) 700 N
Total 17
How to answer it
Forces, Kinematics, Moments and Momentum Study Guide
What this question tests
This comprehensive mechanics question assesses your ability to apply equations of motion (SUVAT), Newton's laws of motion, moments in equilibrium, graphical data interpretation, and power relationships ($P = Fv$). You will need to carefully link vector forces, calculate masses from weights, use graphical trends, and apply momentum principles to fluid dynamics (air resistance).
Calculating Stopping Distance
💡 Key Knowledge
- SUVAT equations require uniform acceleration.
- Identify known variables: initial velocity u = 25 m s⁻¹ , final velocity v = 0 m s⁻¹ , acceleration a = -2.8 m s⁻² (negative because it's deceleration).
📐 Calculation Steps
- Select the correct equation omitting time: v² = u² + 2as
- Rearrange for displacement: s = (v² - u²) / (2a)
- Substitute values: s = (0² - 25²) / (2 × -2.8)
- Calculate final value: s = 112 m (or 110 m to 2 sf).
❌ Common Errors
Students often forget to include the negative sign for acceleration or velocity squared manipulation, yielding negative distances. Ensure your signs align consistently.
✅ Correct Answer & Marks
Answer: 112 m (or 110 m )
Determining Initial Horizontal Force
💡 Key Knowledge
- Use Newton's Second Law: F = ma .
- Convert weight to mass using W = mg , where g ≈ 9.81 m s⁻² or 9.8 m s⁻² .
📐 Calculation Steps
- Find trailer mass: m = W / g = 2500 / 9.81 = 254.8 kg
- Apply F = ma with acceleration a = 1.5 m s⁻² : P_H = 254.8 × 1.5
- Calculate force: P_H = 380 N (using g = 9.8 gives 382.6 N , which rounds to 380 N ).
✅ Correct Answer & Marks
Answer: P_H = 380 N
Finding Constant Velocity from Resultant Force
🧠 Exam Technique
When given perpendicular force components and their resultant, use Pythagoras' theorem or trigonometry. Then, transition to graphical data interpretation using the curve provided.
📐 Calculation Steps
- Use Pythagoras to find P_H : P_H = √(P_Res² - P_V²) = √(912² - 762²) = √(831734 - 580644) = √251090 ≈ 501 N (or use trigonometry).
- Locate D ≈ 500 N on the y-axis of Figure 9 (since air resistance equals the horizontal driving force at constant velocity).
- Read across to the curve and down to the x-axis to find v_1 .
✅ Correct Answer & Marks
Answer: v_1 = 10 m s⁻¹ (accept range 9.75 to 10.25 m s⁻¹ )
Line of Action and Moments
💡 Key Knowledge
The moment of a force is defined as moment = force × perpendicular distance from the pivot . If a force's line of action passes directly through the pivot, the perpendicular distance is zero.
✅ Correct Answer & Marks
Answer: The line of action of P_H passes directly through the centre of the trailer's wheel (the pivot), making the perpendicular distance zero ( d = 0 ).
Calculating Higher Velocity
🧠 Exam Technique
Apply the Principle of Moments (sum of clockwise moments = sum of anticlockwise moments) about the trailer's wheel to find the unknown air resistance D , then use the graph.
📐 Calculation Steps
- Set up moments about the wheel: Anticlockwise moment ( D × 0.95 ) = Clockwise moment ( 2500 × 0.80 ), noting P_V = 0 .
- Solve for D : D = (2500 × 0.80) / 0.95 = 2105 N ≈ 2100 N .
- Read value from Figure 9 graph: find D = 2100 N on y-axis and read corresponding velocity v_2 ≈ 20.5 m s⁻¹ .
✅ Correct Answer & Marks
Answer: v_2 = 20.5 m s⁻¹ (accept range 20 to 21 m s⁻¹ )
Explaining Air Resistance via Momentum
💡 Key Knowledge
- Newton's 2nd Law: Force is equal to the rate of change of momentum ( F = Δp / Δt ).
- Newton's 3rd Law: When the trailer exerts a force on the air, the air exerts an equal and opposite force back on the trailer ( D ).
✅ Model Explanation & Marks
As velocity v increases, more air particles are displaced per second, and each particle is given greater momentum. By Newton's second law, a greater rate of change of momentum of the air requires a larger force exerted by the trailer on the air. By Newton's third law, the air exerts an equal and opposite reaction force (air resistance D ) on the trailer, hence D increases.
Calculating Car Air Resistance
🧠 Exam Technique
Use the power equation P = Fv and resolve forces horizontally for the car-trailer system moving at a constant velocity.
📐 Calculation Steps
- Calculate total forward thrust produced by the car's engine at max power: P = F_total × v ⇒ 95 × 10³ W = F_total × 25 m s⁻¹
- Calculate total thrust force: F_total = (95 × 10³) / 25 = 3800 N
- Identify resistive forces opposing the system: trailer air resistance ( D = 3100 N ) and car air resistance ( R_car ).
- Apply equilibrium condition for constant velocity ( Forward Thrust = Total Resistive Forces ): 3800 = 3100 + R_car
- Solve for car air resistance: R_car = 3800 - 3100 = 700 N
✅ Correct Answer & Marks
Answer: Air resistance on car = 700 N
Topics
Physics · 3.4 Mechanics and materials
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.