| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Constant speed up/down incline |
| Difficulty | Standard +0.3 This is a standard M2 work-energy question requiring routine application of power = force × velocity, work-energy principle, and Newton's second law. All parts follow predictable patterns with straightforward calculations and no novel problem-solving insight required. The multi-part structure and 14 marks indicate moderate length, but each step uses familiar M2 techniques, making it slightly easier than the typical A-level question. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| \((\checkmark): F = 20 + 64g \sin \alpha\) | M1 | |
| \(= 64.8 \text{ N}\) | A1 | |
| \(P = Fv = 64.8 \times 5 = 324 \text{ W}\) | M1 A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \((\checkmark): 64g \sin \alpha - 20 = 64a\) | M1 A1 | |
| \(a = 0.3875 \text{ m s}^{-2}\) | A1 | |
| \(v^2 = 5^2 + 2 \times 0.3875 \times 80\) | M1 | |
| \(v = \sqrt{87} = 9.3 \text{ m s}^{-1}\) (2 sf) | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{5}{8} \times 20 = 32 \text{ N}\) | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = \frac{200}{8}\) | B1 | |
| \(\frac{200}{8} + 64g \sin \alpha - 32 = 64a\) | M1 A1 | |
| \(a = 0.59 \text{ m s}^{-2}\) (2 sf) | A1 | (4) |
## Part (a)
$(\checkmark): F = 20 + 64g \sin \alpha$ | M1
$= 64.8 \text{ N}$ | A1
$P = Fv = 64.8 \times 5 = 324 \text{ W}$ | M1 A1 | (4)
## Part (b)
$(\checkmark): 64g \sin \alpha - 20 = 64a$ | M1 A1
$a = 0.3875 \text{ m s}^{-2}$ | A1
$v^2 = 5^2 + 2 \times 0.3875 \times 80$ | M1
$v = \sqrt{87} = 9.3 \text{ m s}^{-1}$ (2 sf) | A1 | (5)
## Part (c)
$\frac{5}{8} \times 20 = 32 \text{ N}$ | B1 | (1)
## Part (d)
$F = \frac{200}{8}$ | B1
$\frac{200}{8} + 64g \sin \alpha - 32 = 64a$ | M1 A1
$a = 0.59 \text{ m s}^{-2}$ (2 sf) | A1 | (4)
**Total: (14 marks)**
---A girl and her bicycle have a combined mass of 64 kg. She cycles up a straight stretch of road which is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac{1}{14}$. She cycles at a constant speed of 5 m s$^{-1}$. When she is cycling at this speed, the resistance to motion from non-gravitational forces has magnitude 20 N.
\begin{enumerate}[label=(\alph*)]
\item Find the rate at which the cyclist is working.
[4]
\end{enumerate}
She now turns round and comes down the same road. Her initial speed is 5 m s$^{-1}$, and the resistance to motion is modelled as remaining constant with magnitude 20 N. She free-wheels down the road for a distance of 80 m. Using this model,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the speed of the cyclist when she has travelled a distance of 80 m.
[5]
\end{enumerate}
The cyclist again moves down the same road, but this time she pedals down the road. The resistance is now modelled as having magnitude proportional to the speed of the cyclist. Her initial speed is again 5 m s$^{-1}$ when the resistance to motion has magnitude 20 N.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the magnitude of the resistance to motion when the speed of the cyclist is 8 m s$^{-1}$.
[1]
\end{enumerate}
The cyclist works at a constant rate of 200 W.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the magnitude of her acceleration when her speed is 8 m s$^{-1}$.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2003 Q6 [14]}}