| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done against friction/resistance - inclined plane or slope |
| Difficulty | Standard +0.3 This is a straightforward work-energy question requiring standard applications of KE = ½mv², PE = mgh, work-energy principle, and P = Fv. All three parts follow directly from formula application with no novel problem-solving insight needed. The multi-part structure and calculation steps place it slightly above average difficulty, but it remains a routine textbook exercise. |
| Spec | 6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02l Power and velocity: P = Fv |
| Answer | Marks |
|---|---|
| Gain in KE is 3240 J | B1 |
| Loss in PE is 9070 J | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Work done is 5830 J | B1ft | [3] ft WD = loss of PE − gain in KE (subject to loss of PE ≠ gain in KE) |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = 5830/250 (= 23.3)\) | B1ft | |
| \([23.3d = \frac{1}{2}80(9^2 - 5^2) \text{ or}\) | M1 | For using WD = Loss of KE or for using \(−R = 80a\) and \(v^2 = u^2 + 2as\) |
| \(−23.3 = 80a \text{ and } 5^2 = 9^2 + 2(−23.3/80)d]\) | ||
| \(d = 96.0\) | A1ft | [3] Accept 96 or 96.1; ft 560000/WD(i) or 2240/R |
| Answer | Marks | Guidance |
|---|---|---|
| Driving force = 425/5 | B1 | |
| \([DF - 23.3 = 80a]\) | M1 | For using Newton's second law |
| Acceleration is 0.771 ms⁻² | A1 | [3] |
**(i)**
Gain in KE is 3240 J | B1 |
Loss in PE is 9070 J | B1 |
**SR (max 1 out of 2) for answers −3240 and −9070**
Work done is 5830 J | B1ft | [3] ft WD = loss of PE − gain in KE (subject to loss of PE ≠ gain in KE)
**(ii)**
$R = 5830/250 (= 23.3)$ | B1ft |
$[23.3d = \frac{1}{2}80(9^2 - 5^2) \text{ or}$ | M1 | For using WD = Loss of KE or for using $−R = 80a$ and $v^2 = u^2 + 2as$
$−23.3 = 80a \text{ and } 5^2 = 9^2 + 2(−23.3/80)d]$ | |
$d = 96.0$ | A1ft | [3] Accept 96 or 96.1; ft 560000/WD(i) or 2240/R
**(iii)**
Driving force = 425/5 | B1 |
$[DF - 23.3 = 80a]$ | M1 | For using Newton's second law
Acceleration is 0.771 ms⁻² | A1 | [3]\includegraphics{figure_5}
A cyclist and his machine have a total mass of 80 kg. The cyclist starts from rest at the top $A$ of a straight path and freewheels (moves without pedalling or braking) down the path to $B$. The path $AB$ is inclined at 2.6° to the horizontal and is of length 250 m (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Given that the cyclist passes through $B$ with speed 9 m s$^{-1}$, find the gain in kinetic energy and the loss in potential energy of the cyclist and his machine. Hence find the work done against the resistance to motion of the cyclist and his machine. [3]
\end{enumerate}
The cyclist continues to freewheel along a horizontal straight path $BD$ until he reaches the point $C$, where the distance $BC$ is $d$ m. His speed at $C$ is 5 m s$^{-1}$. The resistance to motion is constant, and is the same on $BD$ as on $AB$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the value of $d$. [3]
\end{enumerate}
The cyclist starts to pedal at $C$, generating 425 W of power.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the acceleration of the cyclist immediately after passing through $C$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2009 Q5 [9]}}