| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find steady/maximum speed given power |
| Difficulty | Moderate -0.8 This is a straightforward application of the power equation P = Fv with constant speed (so forces balance). Part (a) requires one equation, part (b) adds a simple component for weight down the slope. Both parts are routine calculations with no problem-solving insight needed, making it easier than average. |
| Spec | 6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Driving force \(= \frac{P}{v}\) | B1 | |
| \(\frac{21000}{v} = 600 \Rightarrow v = 35 \text{ ms}^{-1}\) | M1 A1 | |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{P}{v} = 600 + 1200g \cdot \frac{1}{14}\) | M1 A1 | |
| \((= 1440 \text{ N})\) | ||
| \(\frac{21000}{v} = 1440 \Rightarrow v = \frac{21000}{1440} \approx 14.6 \text{ or } 15 \text{ ms}^{-1}\) | M1 A1 | |
| (4) |
## Question 1:
### Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| Driving force $= \frac{P}{v}$ | B1 | |
| $\frac{21000}{v} = 600 \Rightarrow v = 35 \text{ ms}^{-1}$ | M1 A1 | |
| | **(3)** | |
### Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{P}{v} = 600 + 1200g \cdot \frac{1}{14}$ | M1 A1 | |
| $(= 1440 \text{ N})$ | | |
| $\frac{21000}{v} = 1440 \Rightarrow v = \frac{21000}{1440} \approx 14.6 \text{ or } 15 \text{ ms}^{-1}$ | M1 A1 | |
| | **(4)** | |
---\begin{enumerate}
\item A car of mass 1200 kg moves along a straight horizontal road. The resistance to motion of the car from non-gravitational forces is of constant magnitude 600 N . The car moves with constant speed and the engine of the car is working at a rate of 21 kW .\\
(a) Find the speed of the car.
\end{enumerate}
The car moves up a hill inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 14 }$.\\
The car's engine continues to work at 21 kW , and the resistance to motion from nongravitational forces remains of magnitude 600 N .\\
(b) Find the constant speed at which the car can move up the hill.\\
\hfill \mbox{\textit{Edexcel M2 2005 Q1 [7]}}