| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Force depends on velocity v |
| Difficulty | Challenging +1.2 This is a multi-part variable force mechanics problem requiring force resolution, differential equations, and consideration of direction changes. Part (i) is standard force resolution leading to a separable DE (showing a given result). Parts (ii) and (iii) require solving the DE and handling the motion reversal, which is moderately challenging but follows standard M2 techniques without requiring novel insight. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)3.03v Motion on rough surface: including inclined planes6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (i) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.2gcos30°/(2 3) | A1 | dv/dt = –2.5v – 5 – (5 3)/(2 3) |
| dv/dt = –2.5(3 + v) AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) ∫dv/(3+v)=−2.5 ∫dt | M1 | Separates variables and integrates |
| ln(3 + v) = –2.5t (+ c) | A1 | |
| t = 0, v = 2, hence c = ln5 | Or equivalent use of limits | |
| ln3 = 2.5T + ln5 | M1 | 0 T |
| Answer | Marks |
|---|---|
| T = 0.204 | A1 |
| [4] | T = 0.4ln(5/3) |
| Answer | Marks | Guidance |
|---|---|---|
| 0.5v | M1 | dv/dt = 5 – 2.5v – (5 3)/(2 3) |
| ∫dv/(1 -v) = 2.5 ∫dt | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| t = 0, v = 0, hence c = 0 | B1 | Or equivalent |
| –ln(1 – v) = 2.5x0.4ln(5/3) | M1 | Uses t = T |
| Answer | Marks |
|---|---|
| v = 0.4ms | A1 |
Question 6:
6 | (i) | M1 | N2L with 3 force terms
0.2dv/dt = –0.5v – 0.2gsin30° –
0.2gcos30°/(2 3) | A1 | dv/dt = –2.5v – 5 – (5 3)/(2 3)
dv/dt = –2.5(3 + v) AG | A1
[3]
(ii) ∫dv/(3+v)=−2.5 ∫dt | M1 | Separates variables and integrates
ln(3 + v) = –2.5t (+ c) | A1
t = 0, v = 2, hence c = ln5 | Or equivalent use of limits
ln3 = 2.5T + ln5 | M1 | 0 T
[ln(3 + v)] = [–2.5]
2 0
T = 0.204 | A1
[4] | T = 0.4ln(5/3)
(iii) 0.2dv/dt = 0.2gsin30° – 0.2gcos30°/(2 3) –
0.5v | M1 | dv/dt = 5 – 2.5v – (5 3)/(2 3)
∫dv/(1 -v) = 2.5 ∫dt | A1
–ln(1 – v) = 2.5t (+ c)
t = 0, v = 0, hence c = 0 | B1 | Or equivalent
–ln(1 – v) = 2.5x0.4ln(5/3) | M1 | Uses t = T
–1
v = 0.4ms | A1
[5]\includegraphics{figure_6}
A particle $P$ of mass $0.2$ kg is projected with velocity $2$ m s$^{-1}$ upwards along a line of greatest slope on a plane inclined at $30°$ to the horizontal (see diagram). Air resistance of magnitude $0.5v$ N opposes the motion of $P$, where $v$ m s$^{-1}$ is the velocity of $P$ at time $t$ s after projection. The coefficient of friction between $P$ and the plane is $\frac{1}{2\sqrt{3}}$. The particle $P$ reaches a position of instantaneous rest when $t = T$.
\begin{enumerate}[label=(\roman*)]
\item Show that, while $P$ is moving up the plane, $\frac{dv}{dt} = -2.5(3 + v)$. [3]
\item Calculate $T$. [4]
\item Calculate the speed of $P$ when $t = 2T$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2010 Q6 [12]}}