A-Level · AS · extended response
A braking model tested at two speeds
AS Level foundations · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Resisting force on the rough strip
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Resisting force / N
Horizontal axis (x): Distance along strip / m
Read the exact data as a table
| Distance along strip / m | Resisting force / N |
|---|---|
| 0 | 4 |
| 4 | 8 |
| 6 | 8 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
A 2.0 kg trolley reaches a level rough strip after descending from rest through height 1.8 m on a smooth ramp. Use g = 10 m s⁻². On the strip the opposing force depends on distance as shown. It remains 8 N beyond x = 4 m. Neglect rotational energy. Supplied data (Distance along strip / m; Resisting force / N): (0, 4), (4, 8), (6, 8).
- (a)
Find entry speed and stopping distance, using the force–distance graph rather than a constant deceleration formula.
3 marks - (b)
If the entry speed doubles, determine the new stopping distance and explain the energy scaling.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Find entry speed and stopping distance, using the force–distance graph rather than a constant deceleration formula.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
The trolley gains mgh = 36 J, so ½(2)v² = 36 and v = 6.0 m s⁻¹.
- 2
Work against resistance over the first 4 m is trapezium area ½(4+8)(4) = 24 J, leaving 12 J of kinetic energy.
- 3
At 8 N it travels another 12/8 = 1.5 m, stopping at 5.5 m. Resistance is not constant on the first section, so neither is deceleration.
(b)
If the entry speed doubles, determine the new stopping distance and explain the energy scaling.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Kinetic energy varies as v², so doubling speed multiplies it by four, giving 144 J.
- 2
The first 4 m still dissipates 24 J. The remaining 120 J requires 120/8 = 15 m beyond that point.
- 3
Total distance is 19 m. It is not four times 5.5 m because the spatial resistance profile has a lower-force entry section.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- The trolley gains mgh = 36 J, so ½(2)v² = 36 and v = 6.0 m s⁻¹.
- Work against resistance over the first 4 m is trapezium area ½(4+8)(4) = 24 J, leaving 12 J of kinetic energy.
- At 8 N it travels another 12/8 = 1.5 m, stopping at 5.5 m. Resistance is not constant on the first section, so neither is deceleration.
- Kinetic energy varies as v², so doubling speed multiplies it by four, giving 144 J.
- The first 4 m still dissipates 24 J. The remaining 120 J requires 120/8 = 15 m beyond that point.
- Total distance is 19 m. It is not four times 5.5 m because the spatial resistance profile has a lower-force entry section.
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