| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - horizontal motion or engine power |
| Difficulty | Standard +0.3 This is a standard M2 mechanics question on variable force with resistance proportional to v². Part (a) requires applying F=ma with power P=Fv (routine formula manipulation). Part (b)(i) is trivial (engine off means no driving force). Part (b)(ii) involves separating variables and integrating 1/v² - a standard technique practiced extensively in M2. All steps are predictable textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| \(600\frac{dv}{dt} - \frac{8000}{v} + kv^2 = 0\) | M1A1 m1 A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(-kv^2 = 600\frac{dv}{dt}\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore t = \frac{30}{k}\) | M1 A1 A1 M1 A1 | 5 marks |
**(a)** Power of engine is 8kW
$\therefore$ Force exerted by engine = $\frac{8000}{v}$
Using F = ma:
$\frac{8000}{v} - kv^2 = 600\frac{dv}{dt}$
$600\frac{dv}{dt} - \frac{8000}{v} + kv^2 = 0$ | M1A1 m1 A1 | 4 marks | AG
**(b)(i)** When engine is turned off, power is zero:
$-kv^2 = 600\frac{dv}{dt}$ | B1 | 1 mark | AG
**(ii)** $\int 600\frac{dv}{v^2} = -\int k \, dt$
$-\frac{600}{v} = -kt + c$
When $t = 0$, $v = 20$:
$c = -\frac{600}{20} = -30$
$\therefore \frac{600}{v} = kt + 30$
When $v = 10$, $kt = 30$:
$\therefore t = \frac{30}{k}$ | M1 A1 A1 M1 A1 | 5 marks | SC3 for $\frac{30}{k}$
**Total: 10 marks**
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**GRAND TOTAL: 75 marks**8 A car of mass 600 kg is driven along a straight horizontal road. The resistance to motion of the car is $k v ^ { 2 }$ newtons, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the velocity of the car at time $t$ seconds and $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item When the engine of the car has power 8 kW , show that the equation of motion of the car is
$$600 \frac { \mathrm {~d} v } { \mathrm {~d} t } - \frac { 8000 } { v } + k v ^ { 2 } = 0$$
\item When the velocity of the car is $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the engine is turned off.
\begin{enumerate}[label=(\roman*)]
\item Show that the equation of motion of the car now becomes
$$600 \frac { \mathrm {~d} v } { \mathrm {~d} t } = - k v ^ { 2 }$$
\item Find, in terms of $k$, the time taken for the velocity of the car to drop to $10 \mathrm {~ms} ^ { - 1 }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M2 2008 Q8 [10]}}