| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on smooth curved surface |
| Difficulty | Moderate -0.3 This is a straightforward application of the work-energy principle with standard values. Part (a) uses conservation of energy with no resistance (a routine calculation), and part (b) adds work against resistance but still follows a direct formula application. The question requires minimal problem-solving insight and uses clean numbers designed for easy calculation. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Initial \(KE = \frac{1}{2} \times 0.2 \times 5^2\) or Final \(KE = \frac{1}{2} \times 0.2 \times 3^2\) | B1 | |
| \(\frac{1}{2} \times 0.2 \times 5^2 = 0.2gh + \frac{1}{2} \times 0.2 \times 3^2\) | M1 | Use conservation of energy |
| \(h = 0.8\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Apply work-energy equation from \(A\) to \(C\) | M1 | |
| \(\frac{1}{2} \times 0.2 \times 5^2 - 3.1 + 0.2g \times 0.5 = \frac{1}{2} \times 0.2v^2\) | A1 | Correct work-energy equation |
| Speed \(= 2\ \text{ms}^{-1}\) | A1 | |
| Total: 3 |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Initial $KE = \frac{1}{2} \times 0.2 \times 5^2$ or Final $KE = \frac{1}{2} \times 0.2 \times 3^2$ | B1 | |
| $\frac{1}{2} \times 0.2 \times 5^2 = 0.2gh + \frac{1}{2} \times 0.2 \times 3^2$ | M1 | Use conservation of energy |
| $h = 0.8$ | A1 | |
| **Total: 3** | | |
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Apply work-energy equation from $A$ to $C$ | M1 | |
| $\frac{1}{2} \times 0.2 \times 5^2 - 3.1 + 0.2g \times 0.5 = \frac{1}{2} \times 0.2v^2$ | A1 | Correct work-energy equation |
| Speed $= 2\ \text{ms}^{-1}$ | A1 | |
| **Total: 3** | | |3\\
\includegraphics[max width=\textwidth, alt={}, center]{9ac08732-e825-473a-943c-8ad8e9e0bc17-04_519_1018_260_561}
The diagram shows the vertical cross-section of a surface. $A , B$ and $C$ are three points on the crosssection. The level of $B$ is $h \mathrm {~m}$ above the level of $A$. The level of $C$ is 0.5 m below the level of $A$. A particle of mass 0.2 kg is projected up the slope from $A$ with initial speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The particle remains in contact with the surface as it travels from $A$ to $C$.
\begin{enumerate}[label=(\alph*)]
\item Given that the particle reaches $B$ with a speed of $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and that there is no resistance force, find $h$.
\item It is given instead that there is a resistance force and that the particle does 3.1 J of work against the resistance force as it travels from $A$ to $C$.
Find the speed of the particle when it reaches $C$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q3 [6]}}