| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find power at constant speed |
| Difficulty | Moderate -0.3 This is a straightforward M2 work-energy-power question requiring standard application of P=Fv and F=ma. Part (a) uses Newton's second law to find resistance, part (b) applies power formula with component of weight on incline. All steps are routine with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 3.03f Weight: W=mg6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Diagram showing forces with \(0.4 \text{ ms}^{-2}\) acceleration | B1 | |
| \(T = \frac{30000}{20} = 1500\) | M1 | Use of \(P = Fv\) |
| \(T - R = 1800a\) | A1 | Equation of motion. Need all 3 terms. Condone sign errors. Equation correct (their \(T\)) |
| \(T - R = 1800 \times 0.4\), \(R = 1500 - 1800 \times 0.4 = 780\) | A1 | Only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Diagram showing forces on incline | M1 | |
| \(T - 1800g\sin\alpha - R = 0\) | A1 | Equation of motion. Need all 3 terms. Weight must be resolved. Condone cos for sin. Condone sign errors. Correct equation. Allow with \(R\) not substituted or with their \(R\) |
| \(T = 1800 \times \frac{1}{12}g + 780\) | DM1 | |
| \(\text{Power} = \left(1800 \times \frac{1}{12}g + 780\right) \times 20\) | A1 | Use of \(P = Tv\). Correctly substituted equation (for their \(R\)) |
| \(= 45000 \text{ W}\) or \(45 \text{ kW}\) | A1 | cao |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Diagram showing forces with $0.4 \text{ ms}^{-2}$ acceleration | B1 | |
| $T = \frac{30000}{20} = 1500$ | M1 | Use of $P = Fv$ |
| $T - R = 1800a$ | A1 | Equation of motion. Need all 3 terms. Condone sign errors. Equation correct (their $T$) |
| $T - R = 1800 \times 0.4$, $R = 1500 - 1800 \times 0.4 = 780$ | A1 | Only |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Diagram showing forces on incline | M1 | |
| $T - 1800g\sin\alpha - R = 0$ | A1 | Equation of motion. Need all 3 terms. Weight must be resolved. Condone cos for sin. Condone sign errors. Correct equation. Allow with $R$ not substituted or with their $R$ |
| $T = 1800 \times \frac{1}{12}g + 780$ | DM1 | |
| $\text{Power} = \left(1800 \times \frac{1}{12}g + 780\right) \times 20$ | A1 | Use of $P = Tv$. Correctly substituted equation (for their $R$) |
| $= 45000 \text{ W}$ or $45 \text{ kW}$ | A1 | cao |
---2. A lorry of mass 1800 kg travels along a straight horizontal road. The lorry's engine is working at a constant rate of 30 kW . When the lorry's speed is $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, its acceleration is $0.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. The magnitude of the resistance to the motion of the lorry is $R$ newtons.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $R$.
The lorry now travels up a straight road which is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 12 }$. The magnitude of the non-gravitational resistance to motion is $R$ newtons. The lorry travels at a constant speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Find the new rate of working of the lorry's engine.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q2 [9]}}