| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 3 |
| Topic | Variable Force |
| Type | Air resistance kv² - horizontal motion or engine power |
| Difficulty | Standard +0.3 This is a standard mechanics question on variable force with air resistance proportional to v². Parts (i)-(ii) involve straightforward application of P=Fv and equilibrium. Part (iii) requires using F=ma with the chain rule (a = v dv/dx), which is a standard technique. Part (iv) involves integration that requires partial fractions, but the 'show that' format provides the target answer. While multi-step, all techniques are standard for Further Maths mechanics with no novel insight required. |
| Spec | 1.08d Evaluate definite integrals: between limits6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \[= \left[-300\ln | 20000 - v^3 | \right]_{20}^{25}\] A1 |
**Question 5**
**(i)** $R = \dfrac{8800}{20} = 440$ where $R$ is the resistance in newtons M1A1 **[2 marks]**
**(ii)** $440 = k \times 20^2 \Rightarrow k = \dfrac{440}{400} = 1.1$ M1A1 **[2 marks]**
**(iii)** Newton II: $\dfrac{22000}{v} - 1.1v^2 = 990v\dfrac{\mathrm{d}v}{\mathrm{d}x}$ M1A1
$$\Rightarrow \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{990v}\left(\frac{22000}{v} - 1.1v^2\right) = \frac{2200}{99v^2} - \frac{11v}{9900} = \frac{20000 - v^3}{900v^2}$$ (AG) A1 **[3 marks]**
**(iv)** $$x = \int_{20}^{25} \frac{900v^2}{20000 - v^3}\,\mathrm{d}v = -300\int_{20}^{25}\frac{-3v^2}{20000-v^3}\,\mathrm{d}v$$ M1
$$= \left[-300\ln|20000 - v^3|\right]_{20}^{25}$$ A1
$$= -300(\ln 4375 - \ln 12000) = 300\ln\left(\frac{12000}{4375}\right) = 303 \quad \text{(3sf)}$$ M1A1 **[4 marks]**5 When a car of mass 990 kg moves at a constant speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along a horizontal straight road, the power of its engine is 8.8 kW .\\
(i) Find the magnitude of the resistance to the motion of the car at this speed.\\
(ii) Assuming that the resistance has magnitude $k v ^ { 2 }$ newtons when the speed is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find the value of the constant $k$.
The power of the engine is now increased to 22 kW and remains constant at this value.\\
(iii) Using the model in part (ii), show that
$$\frac { \mathrm { d } v } { \mathrm {~d} x } = \frac { 20000 - v ^ { 3 } } { 900 v ^ { 2 } } .$$
(iv) Hence show that the car moves about 300 m as its speed increases from $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q5 [3]}}