| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Topic | Variable Force |
| Type | Air resistance kv² - projected vertically upward |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics problem on variable force with air resistance proportional to v². While it requires setting up and solving a differential equation (separating variables for dv/dt = -g - kv² on ascent), the method is well-established and commonly practiced. The integration involves a standard arctanh form, and both parts follow directly from the setup. It's harder than typical A-level due to the differential equations and Further Maths content, but represents a routine application of the technique rather than requiring novel insight. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)3.02h Motion under gravity: vector form3.03a Force: vector nature and diagrams3.03b Newton's first law: equilibrium6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(H = \dfrac{1}{2k}\left[\ln | g + kv^2 | \right]_0^u\) — A1 (Correct indefinite integral) |
**Question 10(i) and 10(ii)**
**(i)**
$-(mg + mkv^2) = m\dfrac{\mathrm{d}v}{\mathrm{d}t}$ — B1 (Allow $+$ only if $\downarrow$ explicit)
$-\int_0^T \mathrm{d}t = \int_u^0 \dfrac{1}{g + kv^2}\,\mathrm{d}v$ — M1 (Separate and insert integral signs)
$= \dfrac{1}{k}\int_{\frac{g}{k}}^0 \dfrac{1}{\frac{g}{k} + v^2}\,\mathrm{d}v$ — M1 (Or indefinite integral and find $c$)
$T = \dfrac{1}{k}\left[\sqrt{\dfrac{k}{g}}\tan^{-1}\sqrt{\dfrac{k}{g}}\,v\right]_0^u$ — A1 (Correct indefinite integral)
$= \dfrac{1}{\sqrt{gk}}\tan^{-1}\sqrt{\dfrac{k}{g}}\,u$ — A1
**[5]**
**(ii)**
$-(mg + mkv^2) = mv\dfrac{\mathrm{d}v}{\mathrm{d}x}$ — B1 (Allow $+$ only if $\downarrow$ explicit)
$-\int_0^H \mathrm{d}x = \dfrac{1}{2k}\int_u^0 \dfrac{2kv}{g + kv^2}\,\mathrm{d}v$ — M1, M1 (Or indefinite integral and find $c$)
$H = \dfrac{1}{2k}\left[\ln|g + kv^2|\right]_0^u$ — A1 (Correct indefinite integral)
$= \dfrac{1}{2k}\ln\!\left(\dfrac{g + ku^2}{g}\right)$ — A1
**[5]**10 A small body of mass $m$ is thrown vertically upwards with initial velocity $u$. Resistance to motion is $k v ^ { 2 }$ per unit mass, where the velocity is $v$ and $k$ is a positive constant. Find, in terms of $u , g$ and $k$,\\
(i) the time taken to reach the greatest height,\\
(ii) the greatest height to which the body will rise.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2015 Q10 [5]}}