AQA AS Level Physics Paper 1, June 2018: Question 2
9 marks · Medium difficulty · Extended Answer
Calculate the speed of a truck moving down a ramp from rest with a vertical displacement of 8.0 m, discuss its acceleration with a varying slope compared to a straight ramp, explain why the final speed is the same regardless of ramp shape given equal vertical displacement, and discuss momentum changes when rain falls into the moving truck.
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Question text
02.1 Figure 1 shows a truck moving freely down a ramp inclined at an angle to the
horizontal.
Figure 1
The truck starts from rest at the top of the ramp and reaches point A. Friction and air
resistance are negligible.
As the truck moves down the ramp to point A, its centre of mass has a total vertical
displacement of 8.0 m
Calculate the speed of the truck at point A.
[2 marks]
5 m s−1
speed =
02.2 Figure 2 shows the truck moving down a ramp with a varying slope.
Figure 2
The truck starts from rest and moves freely down the ramp. It reaches point C and
then moves along the horizontal runway to D. Friction and air resistance are
negligible.
Discuss how the acceleration of the truck in Figure 2 differs from the acceleration of
the truck in Figure 1.
[3 marks]
02.3 The total vertical displacement of the centre of mass of the truck in Figure 2 is also
8.0 m
The speed of the truck when it reaches the horizontal runway is the same as the
speed of the truck in Figure 1 when it reaches point A.
Explain why.
[1 mark]
02.4 The horizontal runway in Figure 2 has negligible friction and air resistance. As the
truck moves along the runway, it starts to rain. The rain falls vertically and water
collects in the truck.
Discuss whether there are any changes in the momentum of the truck and collected
water.
[3 marks]
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidelines Mark
(use of gain in Ek = loss in Ep) Bald correct answer scores 1 mark
1/2mv2 = mgh 2 2
If use v = u +2as then zero
1/2v2 = 9.81 × 8.0
02.1 Unless resolved g along slope
(v = (2 × 9.81 × 8.0)) = 13 (12.5) (m s−1)
If use 10 for g (-1) 1
Gets second mark if answer rounds to 13 1
THREE FROM:
acceleration of truck in Fig.1 is constant
In Fig.2
acceleration is greater/greatest at start/top
02.2 acceleration decreases
reference to zero acceleration/uniform velocity between C and D
(3 MAX)
because the component of weight/acceleration parallel to the slope
changes
the loss of (gravitational) potential energy is the same
02.3 1
hence gain in kinetic energy is the same
THREE FROM: If say: ‘vertical momentum/velocity of rain
rain has no (initial) horizontal momentum drops/water changes to horizontal
vertical momentum of rainwater decreases (momentum/velocity)’ score 2 marks
there is no external (horizontal) impulse/force on the truck (and water 1
02.4
system) 1
mass (of truck) increases but speed/velocity decreases (3 MAX)
horizontal momentum of water increases (but horizontal momentum of
– – –
truck decreases by same amount)
(so) no change in (horizontal) momentum of truck and collected Cannot score last mark if stated that speed/velocity
water/total momentum of truck does not change
Total 9
How to answer it
Dynamics, Energy Conservation and Momentum
This question assesses core mechanics principles at AQA AS Level: conservation of energy (gravitational potential energy to kinetic energy), varying acceleration on curved versus straight slopes, and the application of the conservation of momentum in open systems where mass is added (rain falling vertically into a moving truck).
Calculating Speed from Vertical Displacement
✅ Correct Answer
13 m s⁻¹ (or 12.5 m s⁻¹ unrounded)
💡 Key Knowledge
- With negligible friction and air resistance, total mechanical energy is conserved.
- The loss in gravitational potential energy (Ep) equals the gain in kinetic energy (Ek).
- Vertical height ( h ) is independent of the slope angle when calculating energy changes.
📐 Step-by-Step Calculation
- Equate energy changes: 0.5 × m × v² = m × g × h
- Cancel mass ( m ) from both sides: 0.5 × v² = g × h
- Substitute values ( g = 9.81 m s⁻² , h = 8.0 m ): 0.5 × v² = 9.81 × 8.0
- Rearrange for v : v = √(2 × 9.81 × 8.0) = 12.525... m s⁻¹
- Round to 2 significant figures to match input data: 13 m s⁻¹
❌ Common Errors & Examiner Tips
Students sometimes unnecessarily try to resolve components of gravity along the inclined plane using trigonometry ( mg sin(θ) ). Because energy is a scalar quantity, only the vertical displacement matters. Using g = 10 m s⁻² instead of 9.81 m s⁻² loses an accuracy mark.
Comparing Accelerations on Straight vs. Curved Ramps
✅ Correct Answer (Any 3 points)
- Acceleration in Figure 1 is constant.
- In Figure 2, acceleration is greatest at the top/start.
- Acceleration decreases as the truck moves down the curved slope in Figure 2.
- There is zero acceleration / uniform velocity between points C and D.
- This occurs because the component of weight parallel to the slope changes as the angle changes.
🧠 Exam Technique
This is a "Discuss" question requiring clear comparative statements. Do not just describe Figure 1 and then Figure 2 separately; explicitly contrast how the acceleration behaves differently over time/position due to the changing gradient.
❌ Common Errors
A frequent error is claiming the truck travels at constant speed down the whole of Figure 2. Students confuse constant velocity with the horizontal section (C to D), failing to recognize that acceleration occurs along the curved ramp section.
Energy Conservation on a Curved Slope
✅ Correct Answer
The loss of gravitational potential energy is the same because the vertical displacement is identical, resulting in the same gain in kinetic energy.
💡 Key Knowledge
Gravitational potential energy depends strictly on vertical height change ( ΔEp = m g Δh ). The shape of the path taken (straight vs. curved) has no effect on total energy conversion when non-conservative forces like friction are absent.
Momentum in an Open System with Rain
✅ Correct Answer
No change in the total horizontal momentum of the truck and collected water.
💡 Key Knowledge
- Rain falls vertically, meaning it possesses zero initial horizontal momentum.
- As water collects, its mass increases while its horizontal velocity matches the truck. Consequently, the truck's speed decreases, but horizontal momentum remains conserved.
- There are no external horizontal forces/impulses acting on the truck-water system.
❌ Common Errors & Top-Level Distinction
Top-level responses clearly distinguish between horizontal and vertical components. Students lose marks if they carelessly state "momentum does not change" without specifying horizontal momentum, or if they incorrectly claim the truck's speed remains constant despite increasing mass.
Topics
Physics · 3.4 Mechanics and materials
Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.