| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| 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 to form the DE, separation of variables to solve it, and integration to find displacement. All steps are routine textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)3.03d Newton's second law: 2D vectors4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(60\,dv/dt = 150 - 12v\) | M1 | Newton's Second Law, 2 force terms |
| \(5\,dv/dt = 12.5 - v\) | A1 | [2 marks] AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int 5\,dv/(12.5 - v) = \int dt\) | M1 | Separates variables and attempts to integrate |
| \(-5\ln(12.5 - v) = t\ (+c)\) | A1 | |
| \(t = 0,\ v = 11.5\) hence \(c = 0\) | M1 | Explicit after integration (or limits) |
| \(v = 12.5 - e^{-0.2t}\) | A1 | [4 marks] AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \int(12.5 - e^{-0.2t})\,dt\) | ||
| \(x = 12.5t - e^{-0.2t}/(-0.2)\ (+c)\) | B1 | Award if \(+c\) omitted |
| \(t = 0,\ x = 0\) so \(c = -5\) | M1 | Or limits |
| \(x = 12.5t + 5e^{-0.2t} - 5\) | A1 | [3 marks] |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $60\,dv/dt = 150 - 12v$ | M1 | Newton's Second Law, 2 force terms |
| $5\,dv/dt = 12.5 - v$ | A1 | **[2 marks]** AG |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 5\,dv/(12.5 - v) = \int dt$ | M1 | Separates variables and attempts to integrate |
| $-5\ln(12.5 - v) = t\ (+c)$ | A1 | |
| $t = 0,\ v = 11.5$ hence $c = 0$ | M1 | Explicit after integration (or limits) |
| $v = 12.5 - e^{-0.2t}$ | A1 | **[4 marks]** AG |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \int(12.5 - e^{-0.2t})\,dt$ | | |
| $x = 12.5t - e^{-0.2t}/(-0.2)\ (+c)$ | B1 | Award if $+c$ omitted |
| $t = 0,\ x = 0$ so $c = -5$ | M1 | Or limits |
| $x = 12.5t + 5e^{-0.2t} - 5$ | A1 | **[3 marks]** |
---6 A cyclist and her bicycle have a total mass of 60 kg . The cyclist rides in a horizontal straight line, and exerts a constant force in the direction of motion of 150 N . The motion is opposed by a resistance of magnitude $12 v \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the cyclist's speed at time $t \mathrm {~s}$ after passing through a fixed point $A$.\\
(i) Show that $5 \frac { \mathrm {~d} v } { \mathrm {~d} t } = 12.5 - v$.\\
(ii) Given that the cyclist passes through $A$ with speed $11.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, solve this differential equation to show that $v = 12.5 - \mathrm { e } ^ { - 0.2 t }$.\\
(iii) Express the displacement of the cyclist from $A$ in terms of $t$.
\hfill \mbox{\textit{CAIE M2 2015 Q6 [9]}}