| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2020 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find k or constants |
| Difficulty | Standard +0.3 This is a standard FM1 work-energy-power question requiring unit conversion, application of P=Fv and F=ma, and finding terminal velocity when acceleration is zero. The multi-step nature and quadratic resistance add some complexity beyond basic mechanics, but the solution path is straightforward and follows a well-practiced template for this topic. |
| Spec | 6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Scheme | Marks | Guidance |
| \(72 \text{ km h}^{-1} = 20 \text{ m s}^{-1}\) | B1 | |
| Use of \(F = \frac{P}{v}\) and using the model | M1 | Follow through the 72 or their \(v\). Allow for 144 or their 144 |
| Equation of motion and using the model to form equation in \(c\) | M1 | Correct no. of terms required |
| \(\frac{50000}{20} - c \times 20^2 = 1000 \times 2.25\) (\(c = \frac{5}{8}\)) | A1ft | Correct unsimplified equation ft on their 20 |
| Equation of motion and using the model | M1 | Correct no. of terms required |
| \(\frac{50000}{40} - c \times 40^2 = 1000a\) | A1ft | Correct equation ft on their 40 and their \(c\) |
| Solve for \(a\) | M1 | Complete method to solve for \(a\) |
| \(0.25 \text{ (m s}^{-2}\text{)}\) | A1 | Cao (Accept \(\frac{1}{4}\)) |
| Equation of motion horizontally and using the model | M1 | Equation with correct no. of terms, correct structure and in terms of \(W\) only. |
| \(\frac{50000}{W} - \frac{5}{8}W^2 = 0\) (max speed is \(W \text{ m s}^{-1}\)) | A1ft | Correct equation, ft on their \(c\) from part (b). |
| Solve for \(W\) and convert to km h\(^{-1}\) (\(W = 43.088...\)) | M1 | Complete method to solve for \(V\) (including clear attempt to convert units) |
| \(V = 155\) (nearest whole number) | A1 | Cao (The Q asks for a whole number) |
| Scheme | Marks | Guidance |
|--------|-------|----------|
| $72 \text{ km h}^{-1} = 20 \text{ m s}^{-1}$ | B1 | |
| Use of $F = \frac{P}{v}$ and using the model | M1 | Follow through the 72 or their $v$. Allow for 144 or their 144 |
| Equation of motion and using the model to form equation in $c$ | M1 | Correct no. of terms required |
| $\frac{50000}{20} - c \times 20^2 = 1000 \times 2.25$ ($c = \frac{5}{8}$) | A1ft | Correct unsimplified equation ft on their 20 |
| Equation of motion and using the model | M1 | Correct no. of terms required |
| $\frac{50000}{40} - c \times 40^2 = 1000a$ | A1ft | Correct equation ft on their 40 and their $c$ |
| Solve for $a$ | M1 | Complete method to solve for $a$ |
| $0.25 \text{ (m s}^{-2}\text{)}$ | A1 | Cao (Accept $\frac{1}{4}$) |
| Equation of motion horizontally and using the model | M1 | Equation with correct no. of terms, correct structure and in terms of $W$ only. |
| $\frac{50000}{W} - \frac{5}{8}W^2 = 0$ (max speed is $W \text{ m s}^{-1}$) | A1ft | Correct equation, ft on their $c$ from part (b). |
| Solve for $W$ and convert to km h$^{-1}$ ($W = 43.088...$) | M1 | Complete method to solve for $V$ (including clear attempt to convert units) |
| $V = 155$ (nearest whole number) | A1 | Cao (The Q asks for a whole number) |\begin{enumerate}
\item A car of mass 1000 kg moves along a straight horizontal road.
\end{enumerate}
In all circumstances, when the speed of the car is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the resistance to the motion of the car is modelled as a force of magnitude $c v ^ { 2 } \mathrm {~N}$, where $c$ is a constant.
The maximum power that can be developed by the engine of the car is 50 kW .\\
At the instant when the speed of the car is $72 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ and the engine is working at its maximum power, the acceleration of the car is $2.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }$\\
(a) Convert $72 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ into $\mathrm { m } \mathrm { s } ^ { - 1 }$\\
(b) Find the acceleration of the car at the instant when the speed of the car is $144 \mathrm { kmh } ^ { - 1 }$ and the engine is working at its maximum power.
The maximum speed of the car when the engine is working at its maximum power is $V \mathrm {~km} \mathrm {~h} ^ { - 1 }$.\\
(c) Find, to the nearest whole number, the value of $V$.
\hfill \mbox{\textit{Edexcel FM1 AS 2020 Q2 [12]}}