| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find constant speed |
| Difficulty | Standard +0.3 This is a standard M2 power-resistance problem requiring P=Fv at maximum speed (constant velocity, zero acceleration). Part (a) involves finding the resistance constant then solving P=Rv for v. Part (b) adds a component for gravitational resistance down the slope. Both parts use routine mechanics techniques with straightforward algebra, making this slightly easier than average for M2. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks |
|---|---|
| \(2400 = 20k\) | M1 A1 M1 A1 |
| \(k = 120\) | |
| \(84000 = v(120v)\) | |
| \(v = 26.5 \, \text{ms}^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P = v(600g + 120v)\) | M1 M1 A1 | |
| \(120v^2 + 5880v - 84000 = 0\) | ||
| \(v^2 + 49v - 700 = 0\) | ||
| \(v = \frac{(-49 + \sqrt{5201})}{2} = 11.6 \, \text{ms}^{-1}\) | M1 A1 A1 | Total: 10 marks |
### (a)
$2400 = 20k$ | M1 A1 M1 A1 |
$k = 120$ | |
$84000 = v(120v)$ | |
$v = 26.5 \, \text{ms}^{-1}$ | |
### (b)
$P = v(600g + 120v)$ | M1 M1 A1 |
$120v^2 + 5880v - 84000 = 0$ | |
$v^2 + 49v - 700 = 0$ | |
$v = \frac{(-49 + \sqrt{5201})}{2} = 11.6 \, \text{ms}^{-1}$ | M1 A1 A1 | **Total: 10 marks**A lorry of mass 4200 kg can develop a maximum power of 84 kW. On any road the lorry experiences a non-gravitational resisting force which is directly proportional to its speed. When the lorry is travelling at 20 ms$^{-1}$ the resisting force has magnitude 2400 N.
Find the maximum speed of the lorry when it is
\begin{enumerate}[label=(\alph*)]
\item travelling on a horizontal road, [4 marks]
\item climbing a hill inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac{1}{7}$. [6 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q5 [10]}}