| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ladder on smooth wall and rough ground |
| Difficulty | Standard +0.3 This is a standard M2 equilibrium problem requiring moments about a point, resolving forces, and applying equilibrium conditions. The setup is straightforward with clearly defined geometry (uniform rod, smooth ground, rough peg), and students simply need to take moments about C and resolve vertically/horizontally. While it requires careful application of multiple techniques, it follows a well-practiced template with no novel insight needed, making it slightly easier than average. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Resistance at max speed \(= 30 \times 48 = 1440\) N | M1 | Using \(F = 30v\) at \(v = 48\) |
| \(P = Fv = 1440 \times 48 = 69120\) W | A1 | Shown (given answer) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Driving force \(= \frac{69120}{40} = 1728\) N | M1 | Using \(P = Fv\) with \(v = 40\) |
| Resistance \(= 30 \times 40 = 1200\) N | B1 | |
| Net force \(= 1728 - 1200 = 528\) N | M1 | |
| \(a = \frac{528}{1200} = 0.44\ \text{ms}^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Component of weight down slope \(= 1200g\sin 3°\) | M1 | |
| \(= 1200 \times 9.8 \times \sin 3° = 615.7...\) N | A1 | |
| At max constant speed, driving force + weight component = resistance | M1 | Equation of motion along slope |
| \(\frac{69120}{v} + 1200g\sin 3° = 30v\) | M1 A1 | Forming equation |
| \(30v^2 - 615.7v - 69120 = 0\) | M1 | Rearranging to solve |
| \(v = \frac{615.7 + \sqrt{615.7^2 + 4(30)(69120)}}{60}\) | ||
| \(v \approx 52.6\ \text{ms}^{-1}\) | A1 |
# Question 6:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Resistance at max speed $= 30 \times 48 = 1440$ N | M1 | Using $F = 30v$ at $v = 48$ |
| $P = Fv = 1440 \times 48 = 69120$ W | A1 | Shown (given answer) |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Driving force $= \frac{69120}{40} = 1728$ N | M1 | Using $P = Fv$ with $v = 40$ |
| Resistance $= 30 \times 40 = 1200$ N | B1 | |
| Net force $= 1728 - 1200 = 528$ N | M1 | |
| $a = \frac{528}{1200} = 0.44\ \text{ms}^{-2}$ | A1 | |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Component of weight down slope $= 1200g\sin 3°$ | M1 | |
| $= 1200 \times 9.8 \times \sin 3° = 615.7...$ N | A1 | |
| At max constant speed, driving force + weight component = resistance | M1 | Equation of motion along slope |
| $\frac{69120}{v} + 1200g\sin 3° = 30v$ | M1 A1 | Forming equation |
| $30v^2 - 615.7v - 69120 = 0$ | M1 | Rearranging to solve |
| $v = \frac{615.7 + \sqrt{615.7^2 + 4(30)(69120)}}{60}$ | | |
| $v \approx 52.6\ \text{ms}^{-1}$ | A1 | |
---6 When a car, of mass 1200 kg , travels at a speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, it experiences a resistance force of magnitude $30 v$ newtons.
The car has a maximum constant speed of $48 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a straight horizontal road.
\begin{enumerate}[label=(\alph*)]
\item Show that the maximum power of the car is 69120 watts.
\item The car is travelling along a straight horizontal road.
Find the maximum possible acceleration of the car when it is travelling at a speed of $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item The car starts to descend a hill on a straight road which is inclined at an angle of $3 ^ { \circ }$ to the horizontal. Find the maximum possible constant speed of the car as it travels on this road down the hill.
\includegraphics[max width=\textwidth, alt={}, center]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-13_2484_1709_223_153}\\
$7 \quad$ A uniform rod $A B$, of length 4 m and mass 6 kg , rests in equilibrium with one end, $A$, on smooth horizontal ground. The rod rests on a rough horizontal peg at the point $C$, where $A C$ is 3 m . The rod is inclined at an angle of $20 ^ { \circ }$ to the horizontal.\\
\includegraphics[max width=\textwidth, alt={}, center]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-14_422_984_447_529}
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2010 Q6 [13]}}