| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Session | Specimen |
| Marks | 4 |
| 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 mechanics problem requiring standard application of P=Fv and F=ma. Part (a) is a 'show that' requiring one calculation at maximum speed (where driving force equals resistance). Part (b) applies P=Fv to find driving force, subtracts resistance, then uses F=ma. All steps are routine for Further Maths mechanics with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = 30 \times 40 = 120\) | M1 | Uses fact that at max speed driving force equals resistance |
| \(P = Fv\) | B1 | States or uses \(P = Fv\) |
| \(P = (30 \times 40) \times 40 = 48000 \text{ W}\) | A1 | Obtains correct value for power |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F - 30 \times 25 = 1200a\) | M1 | Uses resistance model in a three term equation of motion |
| \(F = 1200a + 750\), \(48000 = 25(1200a + 750)\) | A1 | Obtains a correct equation of motion |
| \(a = \frac{1920 - 750}{1200}\) | M1 | Solves 'their' equation of motion for \(a\) |
| \(= 0.975 \text{ m s}^{-2} = 0.98 \text{ m s}^{-2}\) to 2 sf | A1F | Obtains correct acceleration. FT 'their' equation provided both M1 marks awarded |
## Question 6:
### Part 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = 30 \times 40 = 120$ | M1 | Uses fact that at max speed driving force equals resistance |
| $P = Fv$ | B1 | States or uses $P = Fv$ |
| $P = (30 \times 40) \times 40 = 48000 \text{ W}$ | A1 | Obtains correct value for power |
### Part 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F - 30 \times 25 = 1200a$ | M1 | Uses resistance model in a three term equation of motion |
| $F = 1200a + 750$, $48000 = 25(1200a + 750)$ | A1 | Obtains a correct equation of motion |
| $a = \frac{1920 - 750}{1200}$ | M1 | Solves 'their' equation of motion for $a$ |
| $= 0.975 \text{ m s}^{-2} = 0.98 \text{ m s}^{-2}$ to 2 sf | A1F | Obtains correct acceleration. FT 'their' equation provided both M1 marks awarded |
---6 A car, of mass 1200 kg , moves on a straight horizontal road where it has a maximum speed of $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
When the car travels at a speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ it experiences a resistance force which can be modelled as being of magnitude 30 v newtons.
6
\begin{enumerate}[label=(\alph*)]
\item Show that the power output of the car is 48000 W , when it is travelling at its maximum speed.
6
\item Find the maximum acceleration of the car when it is travelling at a speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics Q6 [4]}}