| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Particle motion - velocity/time (dv/dt = f(v,t)) |
| Difficulty | Standard +0.3 This is a standard mechanics differential equation question requiring Newton's second law, separation of variables, and integration. The steps are clearly signposted (show that, solve, find distance), the algebra is straightforward, and the techniques are routine for M2 level. Slightly above average difficulty only due to the multi-step nature and need for careful algebraic manipulation. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(81a = 135 - 9v\) | M1 | |
| \(\frac{9}{15-v}\frac{dv}{dt} = 1\) (AG) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int \frac{1}{15-v}\,dv = \int \frac{1}{9}\,dt\) | M1 | |
| \(-\ln(15-v) = t/9\ (+c)\) | A1 | |
| \(t=0,\ v=0\), hence \(c = -\ln15\) | M1 | |
| \(\ln\!\left(\frac{15}{15-v}\right) = t/9\), \(15e^{-t/9} = 15-v\), \(v = 15(1-e^{-t/9})\) (AG) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x = \int 15(1-e^{-t/9})\,dt\) | M1 | |
| \(x = 15t + 15e^{-t/9}/(1/9)\ (+c)\) | A1 | |
| \(t=0,\ x=0\), hence \(c=-135\); \(x(9) = 15\times9 + 15\times9e^{-9/9} - 135\) | M1 | |
| \(x(9) = 49.7\) m | A1 [4] |
## Question 6:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $81a = 135 - 9v$ | M1 | |
| $\frac{9}{15-v}\frac{dv}{dt} = 1$ (AG) | A1 [2] | |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{15-v}\,dv = \int \frac{1}{9}\,dt$ | M1 | |
| $-\ln(15-v) = t/9\ (+c)$ | A1 | |
| $t=0,\ v=0$, hence $c = -\ln15$ | M1 | |
| $\ln\!\left(\frac{15}{15-v}\right) = t/9$, $15e^{-t/9} = 15-v$, $v = 15(1-e^{-t/9})$ (AG) | A1 [4] | |
### Part (iii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = \int 15(1-e^{-t/9})\,dt$ | M1 | |
| $x = 15t + 15e^{-t/9}/(1/9)\ (+c)$ | A1 | |
| $t=0,\ x=0$, hence $c=-135$; $x(9) = 15\times9 + 15\times9e^{-9/9} - 135$ | M1 | |
| $x(9) = 49.7$ m | A1 [4] | |
---6 A cyclist and his bicycle have a total mass of 81 kg . The cyclist starts from rest and rides in a straight line. The cyclist exerts a constant force of 135 N and the motion is opposed by a resistance of magnitude $9 v \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the cyclist's speed at time $t \mathrm {~s}$ after starting.\\
(i) Show that $\frac { 9 } { 15 - v } \frac { \mathrm {~d} v } { \mathrm {~d} t } = 1$.\\
(ii) Solve this differential equation to show that $v = 15 \left( 1 - \mathrm { e } ^ { - \frac { 1 } { 9 } t } \right)$.\\
(iii) Find the distance travelled by the cyclist in the first 9 s of the motion.
\hfill \mbox{\textit{CAIE M2 2010 Q6 [10]}}