| 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 | Standard +0.3 This is a standard M2 variable force question using the v dv/dx method. Part (i) requires forming F=ma with v dv/dx, separating variables, and integrating—a routine technique for this topic. Part (ii) involves separating v=dx/dt and integrating again, then analyzing the domain. While it requires multiple steps and careful algebra, it follows a well-established procedure with no novel insight required, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
|---|---|---|
| \(0 < e^{-2t} < 1\) for all \(t \to x < 5\) for all \(t\) | \(M1, A1, A1, B1 \text{ft}, B1\) [5] | For using \(v = dx/dt\) and separating variables. \(\text{ft } x = (B/2)(1 - e^{-2t})\). 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 $t \to x < 5$ for all $t$ | $M1, A1, A1, B1 \text{ft}, B1$ [5] | For using $v = dx/dt$ and separating variables. $\text{ft } x = (B/2)(1 - e^{-2t})$. AG
---7 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 $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and moves towards a fixed point $A$ on the line. At time $t \mathrm {~s}$ the displacement of $P$ from $O$ is $x \mathrm {~m}$ and the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. A resistive force of magnitude (5-x) N acts on $P$ in the direction towards $O$.\\
(i) Form a differential equation in $v$ and $x$. By solving this differential equation, show that $v = 10 - 2 x$.\\
(ii) Find $x$ in terms of $t$, and hence show that the particle is always less than 5 m from $O$.
\hfill \mbox{\textit{CAIE M2 2010 Q7 [11]}}