| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find acceleration given power |
| Difficulty | Standard +0.3 This is a straightforward power-force-velocity problem requiring the standard formula P=Fv and Newton's second law. Part (a) is a 'show that' verification requiring one calculation. Part (b) involves setting up F=ma with driving force P/v and resistance 640v, then solving a simple equation. Standard FM1 mechanics with no novel insight required, slightly easier than average A-level. |
| Spec | 6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Equation of motion parallel to road with \(a = 0\) using model | M1 | Correct no. of terms with \(a=0\), condone sign errors. Given answer so step must be seen, allow verbal form or diagram |
| \(F - 16000 = 0\) | A1 | Correct equation |
| \(P = 16000 \times 25\) | M1 | Use of \(P = Fv\). Independent mark, could be first mark seen |
| \(= 400\,000 = 400\) kW | A1* | Obtain given answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of \(\frac{400\,000}{V}\) | M1 | Use of \(P = Fv\) |
| Equation of motion parallel to road using refined model | M1 | Correct no. of terms, condone sign errors. Dimensionally correct |
| \(\frac{400\,000}{V} - 640V = 16000 \times 2.1\) | A1 | Correct unsimplified equation |
| \(2V^2 + 105V - 1250 = 0 \quad (640V^2 + 33600V - 400000 = 0)\) | A1 | Correct 3-term quadratic |
| Solve for \(V\) | M1 | For solving a 3-term quadratic — implied by correct value of \(V\), otherwise explicit method required |
| \(V = 10\) (i.e. speed is \(10\ \text{m s}^{-1}\)) | A1 | \(V = 10\) only |
## Question 1:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Equation of motion parallel to road with $a = 0$ using model | M1 | Correct no. of terms with $a=0$, condone sign errors. Given answer so step must be seen, allow verbal form or diagram |
| $F - 16000 = 0$ | A1 | Correct equation |
| $P = 16000 \times 25$ | M1 | Use of $P = Fv$. Independent mark, could be first mark seen |
| $= 400\,000 = 400$ kW | A1* | Obtain **given answer** from correct working |
**(4 marks)**
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $\frac{400\,000}{V}$ | M1 | Use of $P = Fv$ |
| Equation of motion parallel to road using refined model | M1 | Correct no. of terms, condone sign errors. Dimensionally correct |
| $\frac{400\,000}{V} - 640V = 16000 \times 2.1$ | A1 | Correct unsimplified equation |
| $2V^2 + 105V - 1250 = 0 \quad (640V^2 + 33600V - 400000 = 0)$ | A1 | Correct 3-term quadratic |
| Solve for $V$ | M1 | For solving a 3-term quadratic — implied by correct value of $V$, otherwise explicit method required |
| $V = 10$ (i.e. speed is $10\ \text{m s}^{-1}$) | A1 | $V = 10$ only |
**(6 marks)**
---\begin{enumerate}
\item A lorry of mass 16000 kg moves along a straight horizontal road.
\end{enumerate}
The lorry moves at a constant speed of $25 \mathrm {~ms} ^ { - 1 }$\\
In an initial model for the motion of the lorry, the resistance to the motion of the lorry is modelled as having constant magnitude 16000 N .\\
(a) Show that the engine of the lorry is working at a rate of 400 kW .
The model for the motion of the lorry along the same road is now refined so that when the speed of the lorry along the same road is $V \mathrm {~ms} ^ { - 1 }$, the resistance to the motion of the lorry is modelled as having magnitude 640 V newtons.
Assuming that the engine of the lorry is working at the same rate of 400 kW\\
(b) use the refined model to find the speed of the lorry when it is accelerating at $2.1 \mathrm {~ms} ^ { - 2 }$
\hfill \mbox{\textit{Edexcel FM1 AS 2019 Q1 [10]}}