| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Topic | Power and driving force |
| Type | Variable resistance: find k or constants |
| Difficulty | Standard +0.3 This is a standard power-resistance mechanics problem with straightforward application of P=Fv at terminal velocity, Newton's second law, and separable variables integration. The steps are routine for Further Maths students: finding k from equilibrium condition, forming a differential equation, and integrating (though the integral requires partial fractions). All techniques are well-practiced and the question provides significant scaffolding through parts (i) and (ii). |
| Spec | 6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = -60 \int_3^7 \frac{-2v}{100 - v^2} dv = \left[ -60 \ln | 100 - v^2 | \right]_3^7\) (Limits not required) M1A1 |
**Question 7(i)**
Tractive force = Resistance at maximum speed $\Rightarrow \frac{75}{10} = 10k \Rightarrow k = \frac{3}{4}$. (AG) **B1** [1]
**Question 7(ii)**
$F = ma \Rightarrow \frac{75}{v} - \frac{3}{4}v = 90\frac{dv}{dt} \Rightarrow \frac{25}{v} - \frac{1}{4}v = 30\frac{dv}{dt}$ (AG) (3 terms req. for M1) **M1A1** [2]
**Question 7(iii)**
$\int_0^t dt = \int_3^7 \frac{120v}{100 - v^2} dv$ **M1**
$t = -60 \int_3^7 \frac{-2v}{100 - v^2} dv = \left[ -60 \ln|100 - v^2| \right]_3^7$ (Limits not required) **M1A1**
$= -60\ln 51 + 60\ln 91 = 60\ln\left(\frac{91}{51}\right)$ $(= 34.7)$ seconds. **M1A1** [5] **[8]**7 A cyclist and her machine have a combined mass of 90 kg and she is riding along a straight horizontal road. She is working at a constant power of 75 W . At time $t$ seconds her speed is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the resistance to motion is $k v \mathrm {~N}$, where $k$ is a constant.\\
(i) If the cyclist's maximum steady speed is $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, show that $k = \frac { 3 } { 4 }$.\\
(ii) Use Newton's second law to show that
$$\frac { 25 } { v } - \frac { v } { 4 } = 30 \frac { \mathrm {~d} v } { \mathrm {~d} t } .$$
(iii) Find the time taken for the cyclist to accelerate from a speed of $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to a speed of $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2012 Q7 [8]}}