| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| 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 straightforward Further Mechanics 1 question applying standard power-force-velocity relationships (P=Fv) and Newton's second law. Part (a) requires equating driving force to resistance at constant speed, while part (b) involves resolving forces on an incline and finding acceleration. The calculations are routine with no conceptual challenges beyond applying familiar formulas, making it slightly easier than average for FM1 material. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Equation of motion: \(F = 500 + 7V\) | M1 | Dimensionally correct. Condone sign errors. Must be using \(a = 0\) |
| Use of \(18000 = F \times V\) | M1 | Correct use of \(P = Fv\) |
| \(\Rightarrow \frac{18000}{V} = 500 + 7V\) | A1 | Correct unsimplified equation. Allow with \(F\). Allow with 18K |
| \(\Rightarrow 7V^2 + 500V - 18000 = 0\) | M1 | Form and solve a 3 term quadratic |
| \(V = 26 \quad (26.309...)\) | A1 | 26 or better \((26.309......)\) |
| (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Equation of motion | M1 | Dimensionally correct. All terms required. Condone sign errors and sin/cos confusion. Omission of \(g\) is an accuracy error |
| \(\frac{18000}{15} - (500 + 7\times15) - 900g \times \frac{1}{21} = 900a\) | A1, A1 | Unsimplified equation with at most one error; Correct unsimplified equation. Allow if \(\sin\theta\) not substituted. Allow with 18K |
| \(a = 0.194\ (0.19)\ (\text{ms}^{-2})\) | A1 | 2 sf or 3 sf only, not \(\frac{7}{36}\) |
| (4 marks) |
## Question 1:
### Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Equation of motion: $F = 500 + 7V$ | M1 | Dimensionally correct. Condone sign errors. Must be using $a = 0$ |
| Use of $18000 = F \times V$ | M1 | Correct use of $P = Fv$ |
| $\Rightarrow \frac{18000}{V} = 500 + 7V$ | A1 | Correct unsimplified equation. Allow with $F$. Allow with 18K |
| $\Rightarrow 7V^2 + 500V - 18000 = 0$ | M1 | Form and solve a 3 term quadratic |
| $V = 26 \quad (26.309...)$ | A1 | 26 or better $(26.309......)$ |
| **(5 marks)** | | |
### Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Equation of motion | M1 | Dimensionally correct. All terms required. Condone sign errors and sin/cos confusion. Omission of $g$ is an accuracy error |
| $\frac{18000}{15} - (500 + 7\times15) - 900g \times \frac{1}{21} = 900a$ | A1, A1 | Unsimplified equation with at most one error; Correct unsimplified equation. Allow if $\sin\theta$ not substituted. Allow with 18K |
| $a = 0.194\ (0.19)\ (\text{ms}^{-2})$ | A1 | 2 sf or 3 sf only, not $\frac{7}{36}$ |
| **(4 marks)** | | |
---\begin{enumerate}
\item A van of mass 900 kg is moving along a straight horizontal road.
\end{enumerate}
At the instant when the speed of the van is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the resistance to the motion of the van is modelled as a force of magnitude $( 500 + 7 v ) \mathrm { N }$.
When the engine of the van is working at a constant rate of 18 kW , the van is moving along the road at a constant speed $V \mathrm {~ms} ^ { - 1 }$\\
(a) Find the value of $V$.
Later on, the van is moving up a straight road that is inclined to the horizontal at an angle $\theta$, where $\sin \theta = \frac { 1 } { 21 }$
At the instant when the speed of the van is $v \mathrm {~ms} ^ { - 1 }$, the resistance to the motion of the van from non-gravitational forces is modelled as a force of magnitude $( 500 + 7 v ) \mathrm { N }$.
The engine of the van is again working at a constant rate of 18 kW .\\
(b) Find the acceleration of the van at the instant when $v = 15$
\hfill \mbox{\textit{Edexcel FM1 2021 Q1 [9]}}