| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Topic | Variable Force |
| Type | Air resistance with other powers |
| Difficulty | Challenging +1.2 This is a variable force mechanics problem requiring integration with resistance proportional to v³. Part (i) involves finding the constant of proportionality and integrating using the chain rule (v dv/dx), which is a standard Further Maths technique. Part (ii) requires separating variables and integrating to find time. While it involves multiple steps and careful algebraic manipulation, the methods are well-established for Pre-U/Further Maths students and the question provides significant scaffolding by giving the result to prove in part (i). |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
(i) $v = 4$, $v\frac{dv}{dx} = -2 \Rightarrow k = \frac{1}{16}$ **B1**
$-kv^3 = 2v\frac{dv}{dx}$ **M1**
$\int \frac{1}{16}\,dx = 2\int -v^{-2}\,dv$ **M1**
$\frac{x}{16} = \frac{2}{v} + c$ **A1**
$x = 0, v = 4 \Rightarrow c = -\frac{1}{2}$ **A1**
$\frac{2}{v} = \frac{x}{16} + \frac{1}{2} \Rightarrow \frac{1}{v} = \frac{x+8}{32}$ (AG) **A1** [6]
(ii) $v = \frac{dx}{dt} = \frac{32}{x+8}$ **M1**
$\int_0^t 32\,dt = \int_0^8 (x+8)\,dx$ **M1A1**
$32t = \left[\frac{x^2}{2} + 8x\right]_0^8 = 96$ **A1**
$\Rightarrow t = 3$ [4]9 A particle of mass 2 kg is moving along the $x$-axis, which is horizontal, against a resistive force which is proportional to the cube of the speed of the particle at any instant. At time $t$ seconds the particle's velocity is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and its displacement is $x \mathrm {~m}$. When $t = 0 , x = 0 , v = 4$ and the retardation is $2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Show that
$$\frac { 1 } { v } = \frac { x + 8 } { 32 } .$$
(ii) Find the time taken to cover the first 8 metres.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2013 Q9 [10]}}