| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| 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 Further Mechanics power-speed-resistance question requiring application of P=Fv and F=ma. Part (a) involves straightforward calculation using given values, while part (b) requires equilibrium on an incline with a simple sine component. The resistance model is given explicitly, making this slightly easier than average FM1 questions which might require more problem-solving insight. |
| Spec | 3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(F = \frac{84000}{12}\) | M1 | Allow use of 84 |
| Equation of motion horizontally | M1 | Correct no. of terms, condone sign errors. Allow if they use 84 or 84000 as the driving force |
| \(\frac{84000}{12} - 490 \times 12 = 5000a\) | A1 | Correct equation |
| \(\frac{28}{125}\) or \(0.224\) or \(0.22\) (m s\(^{-2}\)) | A1 | Accept 0.22 |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(D = \frac{84000}{V}\) | M1 | Allow use of 84 |
| Equation of motion parallel to the road: \(D - 490V - 5000g \sin \alpha = 0\) | M1 | Equation in \(V\) only with correct no. of terms, A condone sign errors and sin/cos confusion and omitted \(g\) with \(D\) in terms of \(V\). |
| \(\frac{84000}{V} - 490V - 5000g \sin \alpha = 0\) | A1 | Correct equation in \(V\) only |
| \(V = 10\) only | A1 | cao |
| (4) |
## 2a
| Use of $F = \frac{84000}{12}$ | M1 | Allow use of 84 |
| Equation of motion horizontally | M1 | Correct no. of terms, condone sign errors. Allow if they use 84 or 84000 as the driving force |
| $\frac{84000}{12} - 490 \times 12 = 5000a$ | A1 | Correct equation |
| $\frac{28}{125}$ or $0.224$ or $0.22$ (m s$^{-2}$) | A1 | Accept 0.22 |
| | (4) | |
## 2b
| Use of $D = \frac{84000}{V}$ | M1 | Allow use of 84 |
| Equation of motion parallel to the road: $D - 490V - 5000g \sin \alpha = 0$ | M1 | Equation in $V$ only with correct no. of terms, A condone sign errors and sin/cos confusion and omitted $g$ with $D$ in terms of $V$. |
| $\frac{84000}{V} - 490V - 5000g \sin \alpha = 0$ | A1 | Correct equation in $V$ only |
| $V = 10$ only | A1 | cao |
| | (4) | |
**Total for Question 2: (8 marks)**
---\begin{enumerate}
\item A lorry has mass 5000 kg .
\end{enumerate}
In all circumstances, when the speed of the lorry is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the resistance to motion of the lorry from non-gravitational forces is modelled as having magnitude $490 v$ newtons.
The lorry moves along a straight horizontal road at $12 \mathrm {~ms} ^ { - 1 }$, with its engine working at a constant rate of 84 kW .
Using the model,\\
(a) find the acceleration of the lorry.
Another straight road is inclined to the horizontal at an angle $\alpha$ where $\sin \alpha = \frac { 1 } { 14 }$\\
With its engine again working at a constant rate of 84 kW , the lorry can maintain a constant speed of $V \mathrm {~ms} ^ { - 1 }$ up the road.
Using the model,\\
(b) find the value of $V$.
\hfill \mbox{\textit{Edexcel FM1 AS 2024 Q2 [8]}}