| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Topic | Projectiles |
| Type | Modelling assumptions and limitations |
| Difficulty | Standard +0.3 This is a standard mechanics problem requiring power-force-velocity relationships and work-energy principles. Part (i) involves straightforward equilibrium at constant speed (P=Fv with forces balanced). Part (ii) requires setting up and solving a work-energy equation with variable velocity, which is routine for A-level mechanics students. The calculations are multi-step but follow standard procedures without requiring novel insight. |
| Spec | 6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow \dfrac{x}{10} = -\dfrac{v^2}{2} - 16v - 256\ln | 16 - v | + c\) M1 |
(i) Driving force at speed $v$ is $\dfrac{400000}{v}$ N **B1**
Resistive forces total $20000 + \dfrac{250000 \times 10}{500} = 25000$ (No penalty for $g = 9.8$ here.) **M1A1**
At maximum speed these are equal:
$\dfrac{400000}{v} = 25000 \Rightarrow v = 16$ ms$^{-1}$. (Allow 16.1 from $g = 9.8$) **M1A1** [5]
(ii) Apply Newton II:
$\dfrac{400000}{v} - 25000 = 250000v\dfrac{\text{d}v}{\text{d}x} \Rightarrow \dfrac{16}{v} - 1 = 10v\dfrac{\text{d}v}{\text{d}x}$ **M1A1**
$\Rightarrow \dfrac{1}{10}\int \text{d}x = \int \dfrac{v^2}{16 - v}\text{d}v \Rightarrow \dfrac{1}{10}\int \text{d}x = \int\!\left(-v - 16 + \dfrac{256}{16 - v}\right)\text{d}v$ **M1A1$\checkmark$**
$\Rightarrow \dfrac{x}{10} = -\dfrac{v^2}{2} - 16v - 256\ln|16 - v| + c$ **M1**
$v = 6$ when $x = 0 \Rightarrow c = 114 + 256\ln 10$ (703.5) (acf) **M1A1**
$\Rightarrow \dfrac{x}{10} = 114 + 256\ln 10 - 72 - 192 - 256\ln 4 = 256\ln\!\left(\dfrac{5}{2}\right) - 150$ when $v = 12$. **M1**
$\Rightarrow x = 846$ (AWRT) **A1** [9]12 A train of mass 250 tonnes is ascending an incline of $\sin ^ { - 1 } \left( \frac { 1 } { 500 } \right)$ and working at 400 kW against resistance to motion which may be regarded as a constant force of 20000 N .\\
(i) Find the constant speed, $V$, with which the train can ascend the incline working at this power.\\
(ii) The train begins to ascend the incline at $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the same power and against the same resistance. Find the distance covered in reaching a speed of $\frac { 3 } { 4 } V$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2011 Q12 [9]}}