| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Maximum speed on incline vs horizontal |
| Difficulty | Standard +0.3 This is a standard M2 mechanics question requiring application of the power-force-velocity relationship (P=Fv) and Newton's second law across three straightforward parts. While it involves multiple steps and unit conversions, the techniques are routine for this module with no novel problem-solving required—slightly easier than average due to its predictable structure. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{F}{v} - R = ma\) \(\therefore \frac{90000}{20} - 1800 = 1200a\) | M2 A1 | |
| \(\therefore a = 2.25 \text{ ms}^{-2}\) | A1 | |
| (b) At max. speed, \(a = 0\), \(\frac{F}{v} - R = 0\) \(\therefore \frac{90000}{v} - 1800 = 0\) so \(v = 50 \text{ ms}^{-1}\) | M1 A1 | |
| \(KE = \frac{1}{2} \times 1200 \times 50^2 = 1\,500\,000 \text{ J} = 1500 \text{ kJ}\) | M1 A1 | |
| (c) \(\frac{F}{v} - R - mg\sin\alpha = 0\) \(\therefore \frac{90000}{25} - 1800 - 1200(9.8)\sin\alpha = 0\) | M2 A1 | |
| \(\sin\alpha = \frac{6.5}{9.8}\) \(\therefore \alpha = 8.8°\) (1dp) | M1 A1 | (13 marks) |
**(a)** $\frac{F}{v} - R = ma$ $\therefore \frac{90000}{20} - 1800 = 1200a$ | M2 A1 |
$\therefore a = 2.25 \text{ ms}^{-2}$ | A1 |
**(b)** At max. speed, $a = 0$, $\frac{F}{v} - R = 0$ $\therefore \frac{90000}{v} - 1800 = 0$ so $v = 50 \text{ ms}^{-1}$ | M1 A1 |
$KE = \frac{1}{2} \times 1200 \times 50^2 = 1\,500\,000 \text{ J} = 1500 \text{ kJ}$ | M1 A1 |
**(c)** $\frac{F}{v} - R - mg\sin\alpha = 0$ $\therefore \frac{90000}{25} - 1800 - 1200(9.8)\sin\alpha = 0$ | M2 A1 |
$\sin\alpha = \frac{6.5}{9.8}$ $\therefore \alpha = 8.8°$ (1dp) | M1 A1 | (13 marks)6. The engine of a car of mass 1200 kg is working at a constant rate of 90 kW as the car moves along a straight horizontal road. The resistive forces opposing the motion of the car are constant and of magnitude 1800 N .
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the car when it is travelling at $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Find, in kJ, the kinetic energy of the car when it is travelling at maximum speed.
The car ascends a hill which is straight and makes an angle $\alpha$ with the horizontal. The power output of the engine and the non-gravitational forces opposing the motion remain the same.
Given that the car can attain a maximum speed of $25 \mathrm {~ms} ^ { - 1 }$,
\item find, in degrees correct to one decimal place, the value of $\alpha$.\\
(5 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q6 [13]}}