| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | Specimen |
| Marks | 9 |
| Topic | Variable Force |
| Type | Power-velocity relationship |
| Difficulty | Standard +0.3 This is a standard power-velocity mechanics problem requiring routine application of P=Fv at steady state, Newton's second law with resistance, separation of variables, and integration. While it involves multiple steps and algebraic manipulation, each technique is well-practiced in Further Maths mechanics with no novel insight required—slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)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 |
From Question 7 in the mark scheme:
**(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**10 A cyclist and her bicycle 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) Given that the steady speed at which the cyclist can move is $10 \mathrm {~ms} ^ { - 1 }$, show that $k = \frac { 3 } { 4 }$.\\
(ii) 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 Q10 [9]}}