| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of velocity (separation of variables) |
| Difficulty | Standard +0.3 This is a standard M2 variable acceleration question following a well-established template: apply Newton's second law to get a differential equation, separate variables and integrate, then integrate velocity for distance. The resistance being proportional to velocity (leading to exponential approach to terminal velocity) is a classic setup. All three parts follow routine procedures with no novel problem-solving required, though it does require competent algebraic manipulation and integration skills across multiple steps. |
| Spec | 4.10c Integrating factor: first order equations6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (i) 81a = 135 – 9v | M1 |
| Answer | Marks |
|---|---|
| 15−v | A1 |
| Answer | Marks |
|---|---|
| 15−v 9 | M1 |
| –ln(15 – v) = t/9 (+ c) | A1 |
| t = 0, v = 0, hence c = –ln15 | M1 |
| Answer | Marks |
|---|---|
| v = 15(1 – e ) AG | A1 |
| Answer | Marks |
|---|---|
| (iii) x = ∫ 15(1 – e –t/9 )dt | M1 |
| Answer | Marks |
|---|---|
| x = 15t + 15e /(1/9) (+ c) | A1 |
| Answer | Marks |
|---|---|
| x(9) = 15x9 + 15x9e – 135 | M1 |
| x(9) = 49.7 m | A1 |
Question 6:
6 | (i) 81a = 135 – 9v | M1
9 dv/dt = 1 AG
15−v | A1
[2]
(ii) ∫ 1 dv = ∫1dt
15−v 9 | M1
–ln(15 – v) = t/9 (+ c) | A1
t = 0, v = 0, hence c = –ln15 | M1
ln( 15 ) = t/9
15−v
–t/9
15e = 15 – v
–t/9
v = 15(1 – e ) AG | A1
[4]
(iii) x = ∫ 15(1 – e –t/9 )dt | M1
–t/9
x = 15t + 15e /(1/9) (+ c) | A1
t = 0, x=0, hence c = –135
–9/9
x(9) = 15x9 + 15x9e – 135 | M1
x(9) = 49.7 m | A1
[4]A cyclist and his bicycle have a total mass of $81 \text{ kg}$. The cyclist starts from rest and rides in a straight line. The cyclist exerts a constant force of $135 \text{ N}$ and the motion is opposed by a resistance of magnitude $9v \text{ N}$, where $v \text{ m s}^{-1}$ is the cyclist's speed at time $t \text{ s}$ after starting.
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac{9}{15-v} \frac{dv}{dt} = 1$. [2]
\item Solve this differential equation to show that $v = 15(1-e^{-\frac{t}{9}})$. [4]
\item Find the distance travelled by the cyclist in the first $9 \text{ s}$ of the motion. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2010 Q6 [10]}}