| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Cyclist or runner: find resistance or speed |
| Difficulty | Standard +0.3 This is a standard M2 work-energy-power question requiring P=Fv and resolving forces on an incline. Part (a) is direct substitution, part (b) involves setting up an equation with weight component, and part (c) requires forming and solving a quadratic equation when resistance ∝ v. All techniques are routine for M2 with no novel insight required, making it slightly easier than average. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(P = Fv = 210 \text{ W}\) | M1 A1 | |
| (b) \(210 = v(42 + 84g \sin \alpha)\) | M1 A1 | |
| \(v = \frac{210}{42 + 4g} = 2.59 \text{ ms}^{-1}\) | M1 A1 | |
| (c) \(R = kv\) | M1 | |
| \(42 = 5k\), so \(k = 8.4\) | A1 | |
| \(210 = v(8.4v + 4g)\) | M1 A1 | |
| \(8.4v^2 + 39.2v - 210 = 0\) | ||
| \(3v^2 + 14v - 75 = 0\) | A1 M1 | |
| \(v = \frac{-14 + \sqrt{1096}}{6} = 3.18 \text{ ms}^{-1}\) | A1 M1 A1 | 14 marks |
(a) $P = Fv = 210 \text{ W}$ | M1 A1 |
(b) $210 = v(42 + 84g \sin \alpha)$ | M1 A1 |
$v = \frac{210}{42 + 4g} = 2.59 \text{ ms}^{-1}$ | M1 A1 |
(c) $R = kv$ | M1 |
$42 = 5k$, so $k = 8.4$ | A1 |
$210 = v(8.4v + 4g)$ | M1 A1 |
$8.4v^2 + 39.2v - 210 = 0$ | |
$3v^2 + 14v - 75 = 0$ | A1 M1 |
$v = \frac{-14 + \sqrt{1096}}{6} = 3.18 \text{ ms}^{-1}$ | A1 M1 A1 | 14 marks7. A cyclist is pedalling along a horizontal cycle track at a constant speed of $5 \mathrm {~ms} ^ { - 1 }$. The air resistance opposing her motion has magnitude 42 N . The combined mass of the cyclist and her machine is 84 kg .
\begin{enumerate}[label=(\alph*)]
\item Find the rate, in W , at which the cyclist is working.
The cyclist now starts to ascend a hill inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 21 }$, at a constant speed.\\
She continues to work at the same rate as before, against the same air resistance.
\item Find the constant speed at which she ascends the hill.
In fact the air resistance is not constant, and a revised model takes account of this by assuming that the air resistance is proportional to the cyclist's speed.
\item Use this model to find an improved estimate of the speed at which she ascends the hill, if her rate of working still remains constant.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [14]}}