| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Topic | Variable Force |
| Type | Air resistance kv² - projected vertically upward |
| Difficulty | Challenging +1.2 This is a standard variable force mechanics problem requiring setup of Newton's second law with air resistance (part i) and integration of a separable differential equation (part ii). While it involves Further Maths content, the techniques are routine: showing a given DE is algebraic manipulation, and finding u requires standard separation of variables with a straightforward integral. The problem is slightly above average difficulty due to the Further Maths context and multi-step nature, but follows predictable patterns for this topic. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
Total: 9 marks
**Part (i):**
$0.6v\dfrac{dv}{dx} = -10 \times 0.6 - 0.024v^2$ **M1** Use $v\,dv/dx$ for $a$ and two other terms in $F = ma$
Hence $v\dfrac{dv}{dx} = -10 - 0.04v^2$ **A1** AG, completely correct, signs not fudged
**Part (ii):**
$\displaystyle\int \dfrac{v}{10 + 0.04v^2}\,dv = -\int dx$ **M1** Separate variables correctly (allow a multiplicative constant error) (ignore $\int$ signs)
$\dfrac{1}{0.08}\ln(10 + 0.04v^2) = -x + c$ oe **A1 A1** Correct indefinite integrals, ignore $c$/limits here
$x = 0,\, v = u \Rightarrow c = \dfrac{1}{0.08}\ln(10 + 0.04u^2)$ **M1** Find $c$ in terms of $u$
$x = \dfrac{1}{0.08}\ln\!\left(\dfrac{10 + 0.04u^2}{10 + 0.04v^2}\right)$ **A1** Correct formula for $v$ and $x$, aef
$v = 0,\quad x = \dfrac{1}{0.08}\ln\!\left(1 + 0.004u^2\right)$ **M1** Substitute $v = 0$ and solve for $u$
$x = 50,\quad u = \sqrt{\dfrac{(e^4 - 1)}{0.004}} = 116 \text{ ms}^{-1}$ **A1** awrt $116 \text{ ms}^{-1}$
Total: **9 marks**12 A particle of mass 0.6 kg is projected vertically upwards from horizontal ground, with initial speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The upwards velocity at any instant is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and the displacement is $x \mathrm {~m}$. Air resistance is modelled by a force $0.024 v ^ { 2 } \mathrm {~N}$ acting downwards.\\
(i) Show that $v$ and $x$ satisfy the differential equation
$$v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 10 - 0.04 v ^ { 2 } .$$
(ii) Find the value of $u$ if the maximum height reached is 50 m .
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2017 Q12 [9]}}