A2 June 2024 Q3
3. A car of mass 1000 kg moves in a straight line along a horizontal road at a constant speed of \(72\ \text{km h}^{-1}\)
- The resistance to the motion of the car is modelled as a constant force of magnitude 900 N
The engine of the car is working at a constant rate of \(P\) kW.
Using the model,
The car now travels in a straight line up a road which is inclined to the horizontal at an angle \(\alpha\), where \(\sin\alpha = \dfrac{2}{49}\)
- In a refined model, the resistance to the motion of the car from non-gravitational forces is now modelled as a force of magnitude \(20v\) newtons, where \(v\ \text{m s}^{-1}\) is the speed of the car
At the instant when the engine of the car is working at a constant rate of 30 kW and the car is moving up the road at \(10\ \text{m s}^{-1}\), the acceleration of the car is \(a\ \text{m s}^{-2}\)
Using the refined model,
Later on, when the engine of the car is again working at a constant rate of 30 kW, the car is moving up the road at a constant speed \(U\ \text{m s}^{-1}\)
Using the refined model,
| Scheme | Marks | AO |
|---|---|---|
| Use \(F = \dfrac{1000P}{v}\) where \(v = \dfrac{72000}{3600}\ (= 20)\) | M1 | 3.3 |
| Use equation of motion: \(F - 900 = 0\) to give equation in \(P\) only: Eg \(\dfrac{1000P}{20} = 900\) | M1 | 3.1b |
| \(P = 18\) | A1 | 1.1b |
| (3) |
Notes
M1: Use of \(P = Fv\), condone \(\dfrac{P}{72}\) or \(\dfrac{P}{20}\)
M1: Use equation of motion to give equation in \(P\) only. Condone use of \(P\) instead of \(1000P\) and condone 72 instead of 20 for method mark.
\(\dfrac{1000P}{20} = 900\) or \(\dfrac{P}{20} = 900\)
A1: Cao Allow \(P = 18\,000\) leading to a final answer of \(P = 18\). Ignore units.
| Scheme | Marks | AO |
|---|---|---|
| Use equation of motion for car and power equation to give equation in \(a\) only | M1 | 3.1b |
| \(\dfrac{30000}{10} - 1000g\sin\alpha - 20 \times 10 = 1000a\) o.e. | A1 A1 | 1.1b 1.1b |
| \(a = 2.4\) | A1 | 1.1b |
| (4) |
Notes
NB: Only penalise use of \(g = 9.81\) once per question
M1: Use equation of motion for the car and power equation to give a dimensionally correct equation in \(a\) only.
All required terms present and no extras, resolving only where necessary. Condone \(\pm\) sign errors and cos/sin confusion.
M0 if a term is missing or if weight is not resolved.
A1: Correct equation in \(a\) only with at most one error.
A1: Correct equation in \(a\) only
A1: Answer of 2.4 only
A0 for \(\tfrac{12}{5}\) (when using \(g = 9.8\), answers must be rounded to 2/3sf, not given in exact form)
A0 for use of \(g = 9.81\)
| Scheme | Marks | AO |
|---|---|---|
| \(F = \dfrac{30\,000}{U}\) | M1 | 3.4 |
| Equation of motion for car | M1 | 3.1b |
| \(F - 1000g\sin\alpha - 20U = 0\) | A1 | 1.1b |
| Correct equation in \(U\) only oe Eg
| A1 | 1.1b |
| \(U = 30\) | A1 | 1.1b |
| (5) | ||
| (12 marks) |
Notes
NB: Only penalise use of \(g = 9.81\) once per question
M1: Use of \(P = Fv\), condone incorrect number of zeros
M0 if using speed of 10 from part (b)
M1: Equation of motion for the car: all required terms present and no extras, resolving only where necessary, dimensionally correct. Condone \(\pm\) sign errors and cos/sin confusion.
M0 if a term is missing or if weight is not resolved.
M0 if using resistance of 200 from part (b)
A1: A correct unsimplified equation, \(F\) does not need to be substituted, sin/cos does not need to be substituted.
A1: A correct equation in \(U\) only.
A1: Answer of 30 only