| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find acceleration given power |
| Difficulty | Standard +0.3 This is a standard Further Maths mechanics question on power-resistance relationships requiring multiple straightforward applications of P=Fv and F=ma. Part (a) is a 'show that' requiring finding the resistance constant, parts (b-c) apply the same method twice, and part (d) requires basic interpretation. While it has multiple parts and is Further Maths content, each step follows a routine procedure with no novel insight required, making it slightly easier than average. |
| Spec | 6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resistance \(= kv\); \(600 = 20k\), so \(k = 30\) | M1 (AO 3.3) | Models resistance force using speed and constant of proportionality |
| \(P = 30 \times 48 \times 48 = 69120\) W | R1 (AO 2.1) | Obtains correct maximum power 69120 W using \(P = Fv\); complete argument required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = 1000a + 30 \times 25\) | M1 (AO 1.1a) | Forms three-term equation of motion; condone sign error |
| \(69120 = (1000a + 30 \times 25) \times 25\) | M1 (AO 3.4) | Uses \(P = Fv\) to form equation to find maximum driving force |
| \(a = 2.01 \text{ m s}^{-2}\) | A1 (AO 1.1b) | AWRT \(2 \text{ m s}^{-2}\); FT their constant of proportionality; condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = 1000a + 30 \times 3\) | — | — |
| \(69120 = (1000a + 30 \times 3) \times 3\) | — | — |
| \(a = 23.0 \text{ m s}^{-2}\) | B1 (AO 1.1b) | AWRT \(23 \text{ m s}^{-2}\); FT their constant of proportionality; condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Acceleration very high when speed \(= 3 \text{ m s}^{-1}\) | E1 (AO 3.5a) | Identifies that acceleration in (c) is very large; OE e.g. car will reach maximum speed in about 2 seconds |
| Model only seems valid for higher speeds | E1 (AO 3.5b) | Suggests that the model is not valid at lower speeds |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resistance $= kv$; $600 = 20k$, so $k = 30$ | M1 (AO 3.3) | Models resistance force using speed and constant of proportionality |
| $P = 30 \times 48 \times 48 = 69120$ W | R1 (AO 2.1) | Obtains correct maximum power 69120 W using $P = Fv$; complete argument required |
---
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = 1000a + 30 \times 25$ | M1 (AO 1.1a) | Forms three-term equation of motion; condone sign error |
| $69120 = (1000a + 30 \times 25) \times 25$ | M1 (AO 3.4) | Uses $P = Fv$ to form equation to find maximum driving force |
| $a = 2.01 \text{ m s}^{-2}$ | A1 (AO 1.1b) | AWRT $2 \text{ m s}^{-2}$; FT their constant of proportionality; condone missing units |
---
## Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = 1000a + 30 \times 3$ | — | — |
| $69120 = (1000a + 30 \times 3) \times 3$ | — | — |
| $a = 23.0 \text{ m s}^{-2}$ | B1 (AO 1.1b) | AWRT $23 \text{ m s}^{-2}$; FT their constant of proportionality; condone missing units |
---
## Question 4(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Acceleration very high when speed $= 3 \text{ m s}^{-1}$ | E1 (AO 3.5a) | Identifies that acceleration in (c) is very large; OE e.g. car will reach maximum speed in about 2 seconds |
| Model only seems valid for higher speeds | E1 (AO 3.5b) | Suggests that the model is not valid at lower speeds |
---4 A car has mass 1000 kg and travels on a straight horizontal road.
The maximum speed of the car on this road is $48 \mathrm {~ms} ^ { - 1 }$\\
In a simple model, it is assumed that the car experiences a resistance force that is proportional to its speed.
When the car travels at $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the magnitude of the resistance force is 600 newtons.
4
\begin{enumerate}[label=(\alph*)]
\item Show that the maximum power of the car is 69120 W\\
4
\item Find the maximum acceleration of the car when it is travelling at $25 \mathrm {~ms} ^ { - 1 }$\\
4
\item Find the maximum acceleration of the car when it is travelling at $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
4
\item Comment on the validity of the model in the context of your answers to parts (b) and (c).
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2020 Q4 [8]}}