| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Maximum speed on horizontal road |
| Difficulty | Moderate -0.3 This is a straightforward M2 mechanics question requiring standard application of P=Fv at maximum speed (part a) and work-energy principle (part b). The calculations are direct with no conceptual subtleties—students simply need to recall the formulas and substitute values correctly. Slightly easier than average due to its routine nature and clear structure. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| At maximum speed, driving force \(=\) resistance | M1 | Correct principle |
| \(P = Fv \Rightarrow F = \frac{P}{v} = \frac{800000}{40}\) | M1 | Correct use of \(P=Fv\) |
| \(F = 20000\) N \(= 20\) kN | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Loss in KE \(= \frac{1}{2}(60000)(40^2 - 36^2)\) | M1 | Correct KE expression |
| \(= \frac{1}{2}(60000)(1600-1296) = \frac{1}{2}(60000)(304) = 9120000\) J | A1 | Correct value |
| Work done by resistance \(= 20000 \times d\) | M1 | Work-energy principle |
| \(20000d = 9120000\) | ||
| \(d = 456\) m | A1 | Correct answer |
# Question 6:
## Part (a):
| At maximum speed, driving force $=$ resistance | M1 | Correct principle |
|---|---|---|
| $P = Fv \Rightarrow F = \frac{P}{v} = \frac{800000}{40}$ | M1 | Correct use of $P=Fv$ |
| $F = 20000$ N $= 20$ kN | A1 | Correct answer |
## Part (b):
| Loss in KE $= \frac{1}{2}(60000)(40^2 - 36^2)$ | M1 | Correct KE expression |
|---|---|---|
| $= \frac{1}{2}(60000)(1600-1296) = \frac{1}{2}(60000)(304) = 9120000$ J | A1 | Correct value |
| Work done by resistance $= 20000 \times d$ | M1 | Work-energy principle |
| $20000d = 9120000$ | | |
| $d = 456$ m | A1 | Correct answer |
---6 A train, of mass 60 tonnes, travels on a straight horizontal track. It has a maximum speed of $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when its engine is working at 800 kW .
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the resistance force acting on the train when the train is travelling at its maximum speed.
\item When the train is travelling at $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the power is turned off. Assume that the resistance force is constant and is equal to that found in part (a). Also assume that this resistance force is the only horizontal force acting on the train.
Use an energy method to find how far the train travels when slowing from $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $36 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2009 Q6 [7]}}