| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable force (position x) - find velocity |
| Difficulty | Challenging +1.2 This is a standard variable force mechanics problem requiring formation and solution of a differential equation using v dv/dx = a, followed by integration to find x(t). While it involves multiple steps and calculus techniques, the methods are routine for M2 students and the question provides the answer to verify against, reducing problem-solving demand. The mathematics is straightforward once the correct approach is identified. |
| Spec | 4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(v^2 = 4(x - 5)^2\) Selects correct square root to obtain \(v = 10 - 2x\) | B1 M1 A1 M1 A1 A1 [6] | For using Newton's second law and \(a = v(dv/dx)\). For separating variables and attempting to integrate. For using \(v(0) = 10\). Any correct expression in x. AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(-\frac{1}{2}\ln(10 - 2x) = t(-\frac{1}{2}\ln B)\) \(B = 10\) (or equivalent) \(x = 5(1 - e^{-2t})\) \(0 < e^{-2t} < 1\) for all \(t → x < 5\) for all \(t\) | M1 A1 A1 B1ft B1 [5] | For using \(v = dx/dt\) and separating variables. \(ftx = (B/2)(1 - e^{-2t})\) AG |
## (i)
$[0.25v(dv/dx) = -(5 - x)]$
$[\int vdv = 4\int(x - 5)dx]$
$v^2/2 = 4(x - 5)^2/2 + A)$
$v^2 = 4(x - 5)^2$ Selects correct square root to obtain $v = 10 - 2x$ | B1 M1 A1 M1 A1 A1 [6] | For using Newton's second law and $a = v(dv/dx)$. For separating variables and attempting to integrate. For using $v(0) = 10$. Any correct expression in x. AG
## (ii)
$[\int \frac{dx}{10 - 2x} = \int dt]$
$-\frac{1}{2}\ln(10 - 2x) = t(-\frac{1}{2}\ln B)$ $B = 10$ (or equivalent) $x = 5(1 - e^{-2t})$ $0 < e^{-2t} < 1$ for all $t → x < 5$ for all $t$ | M1 A1 A1 B1ft B1 [5] | For using $v = dx/dt$ and separating variables. $ftx = (B/2)(1 - e^{-2t})$ AG
---A particle $P$ of mass 0.25 kg moves in a straight line on a smooth horizontal surface. $P$ starts at the point $O$ with speed $10 \text{ m s}^{-1}$ and moves towards a fixed point $A$ on the line. At time $t$ s the displacement of $P$ from $O$ is $x$ m and the velocity of $P$ is $v \text{ m s}^{-1}$. A resistive force of magnitude $(5 - x)$ N acts on $P$ in the direction towards $O$.
\begin{enumerate}[label=(\roman*)]
\item Form a differential equation in $v$ and $x$. By solving this differential equation, show that $v = 10 - 2x$. [6]
\item Find $x$ in terms of $t$, and hence show that the particle is always less than 5 m from $O$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2010 Q7 [11]}}