| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Resistance as kv - finding constant k |
| Difficulty | Moderate -0.3 This is a straightforward M2 mechanics question requiring standard techniques: part (a) uses P=Fv at maximum speed (equilibrium), part (b)(i) applies F=ma with resistance force, and part (b)(ii) solves a separable differential equation. All steps are routine applications of well-practiced methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Max speed \(\equiv\) zero acceleration used; \(\frac{72000}{60} = k \times 60\) giving \(k = 20\) | M1, M1, A1 | Implied |
| (b)(i) \(20v = -500\frac{dv}{dt}\) giving \(\frac{dv}{dt} = -\frac{v}{25}\) | M1, A1 | see \(\frac{dv}{dt}, \pm\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(25\ln 10 - 25\ln 20 = -t\) giving \(t = 25\ln 2\) or \(17.3\) or \(-25\ln\frac{1}{2}\) | M1, A1, A1, m1, A1, A1, A1 | M1 separating variables; Alternative: \(25\ln v = -t(+c)\); \(t = 0, v = 20, c = 25\ln 20\); \(t = t, v = 10, 25\ln 10 = -t + 25\ln 20\); \(t = 25\ln 2\) or 17.3 |
**(a)** Max speed $\equiv$ zero acceleration used; $\frac{72000}{60} = k \times 60$ giving $k = 20$ | M1, M1, A1 | Implied | **Total: 3 marks**
**(b)(i)** $20v = -500\frac{dv}{dt}$ giving $\frac{dv}{dt} = -\frac{v}{25}$ | M1, A1 | see $\frac{dv}{dt}, \pm$ | **Total: 2 marks**
**(ii)** $25\int\frac{dv}{v} = -\int dt$ giving $[25\ln v]_{20}^{10} = -[t]_0^t$
$25\ln 10 - 25\ln 20 = -t$ giving $t = 25\ln 2$ or $17.3$ or $-25\ln\frac{1}{2}$ | M1, A1, A1, m1, A1, A1, A1 | M1 separating variables; Alternative: $25\ln v = -t(+c)$; $t = 0, v = 20, c = 25\ln 20$; $t = t, v = 10, 25\ln 10 = -t + 25\ln 20$; $t = 25\ln 2$ or 17.3 | **Total: 6 marks**
**Question 7 Total: 11 marks**
---7 A motorcycle has a maximum power of 72 kilowatts. The motorcycle and its rider are travelling along a straight horizontal road. When they are moving at a speed of $\mathrm { V } \mathrm { m } \mathrm { s } ^ { - 1 }$, they experience a total resistance force of magnitude $k V$ newtons, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item The maximum speed of the motorcycle and its rider is $60 \mathrm {~ms} ^ { - 1 }$.
Show that $k = 20$.
\item When the motorcycle is travelling at $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the rider allows the motorcycle to freewheel so that the only horizontal force acting is the resistance force. When the motorcycle has been freewheeling for $t$ seconds, its speed is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the magnitude of the resistance force is $20 v$ newtons.
The mass of the motorcycle and its rider is 500 kg .
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac { \mathrm { d } v } { \mathrm {~d} t } = - \frac { v } { 25 }$.
\item Hence find the time that it takes for the speed of the motorcycle to reduce from $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(6 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M2 2007 Q7 [11]}}