| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | Specimen |
| Marks | 9 |
| Topic | Power and driving force |
| Type | Variable resistance: find k or constants |
| Difficulty | Standard +0.3 This is a standard power-resistance mechanics problem requiring routine application of P=Fv at terminal velocity, Newton's second law to form a differential equation, and separation of variables for integration. The algebraic manipulation is straightforward and the question structure is highly scaffolded with parts (i) and (ii) guiding students through the setup before the final integration in (iii). |
| 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) M1 A1 |
**(i)** Tractive force = Resistance at maximum speed $\Rightarrow \frac{75}{10} = 10k \Rightarrow k = \frac{3}{4}$ AG **B1**
**(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 required for M1) **M1 A1**
**(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) **M1 A1**
$= -60\ln 51 + 60\ln 91 = 60\ln\left(\frac{91}{51}\right)$ $(= 34.7)$ seconds. **M1 A1**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 $v \mathrm {~ms} ^ { - 1 }$ and the resistance to motion is $k v \mathrm {~N}$, where $k$ is a constant.\\
(i) If the cyclist's maximum steady speed is $10 \mathrm {~ms} ^ { - 1 }$, 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 $3 \mathrm {~ms} ^ { - 1 }$ to a speed of $7 \mathrm {~ms} ^ { - 1 }$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q7 [9]}}