| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find constant speed |
| Difficulty | Moderate -0.3 This is a standard M2 work-energy-power question requiring straightforward application of P=Fv and F=ma. Part (a) is a 'show that' using maximum speed equilibrium, part (b) applies P=Fv then Newton's second law, and part (c) adds a component of weight. All steps are routine textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03b Newton's first law: equilibrium3.03v Motion on rough surface: including inclined planes6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Using power \(=\) force \(\times\) velocity, Power \(= (40 \times 50) \times 50 = 100,000\) watts | M1 A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| When speed is 25, max force exerted is \(\frac{100000}{25} = 4000\) N. Accelerating force is 3000 N. Using \(F = ma\), \(3000 = 1500a\), \(a = 2\) ms\(^{-2}\) | B1 M1 A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| When van is at maximum speed, force against gravity is \(mg\sin 6\) (parallel to slope). Force against gravity and resistance is \(mg\sin 6 + 40v = 1536.6 + 40v\). Speed is maximum when \(1536.6 + 40v = \frac{100000}{v}\), giving \(40v^2 + 1536.6v - 100,000 = 0\). Speed is 34.4 ms\(^{-1}\) | B1 M1 A1 M1 A1 A1 | 6 marks |
**Part (a)**
Using power $=$ force $\times$ velocity, Power $= (40 \times 50) \times 50 = 100,000$ watts | M1 A1 | 2 marks
**Part (b)**
When speed is 25, max force exerted is $\frac{100000}{25} = 4000$ N. Accelerating force is 3000 N. Using $F = ma$, $3000 = 1500a$, $a = 2$ ms$^{-2}$ | B1 M1 A1 | 3 marks | Need 3 terms e.g. '4000' ± 1000 = ma or 2000 ± 1000 = ma. M0 for 1000 = ma
**Part (c)**
When van is at maximum speed, force against gravity is $mg\sin 6$ (parallel to slope). Force against gravity and resistance is $mg\sin 6 + 40v = 1536.6 + 40v$. Speed is maximum when $1536.6 + 40v = \frac{100000}{v}$, giving $40v^2 + 1536.6v - 100,000 = 0$. Speed is 34.4 ms$^{-1}$ | B1 M1 A1 M1 A1 A1 | 6 marks | For 3 terms; $\frac{100000}{v}$ and 1 other term correct. CAO
**Total: 11 marks**
---4 A van, of mass 1500 kg , has a maximum speed of $50 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a straight horizontal road. When the van travels at a speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, it experiences a resistance force of magnitude $40 v$ newtons.
\begin{enumerate}[label=(\alph*)]
\item Show that the maximum power of the van is 100000 watts.
\item The van is travelling along a straight horizontal road.
Find the maximum possible acceleration of the van when its speed is $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item The van starts to climb a hill which is inclined at $6 ^ { \circ }$ to the horizontal. Find the maximum possible constant speed of the van as it travels in a straight line up the hill.\\
(6 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2008 Q4 [11]}}