| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on slope then horizontal |
| Difficulty | Moderate -0.8 This is a straightforward application of conservation of energy and work-energy principles. Part (i) uses basic PE to KE conversion on a smooth surface (standard formula application), and part (ii) requires calculating work done against friction using the work-energy theorem. Both parts are routine textbook exercises requiring only direct formula application with no problem-solving insight or multi-step reasoning. |
| 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 |
|---|---|---|
| (i) \([\frac{1}{2}v^2 = 10x \times 1.8]\) Speed is \(6\text{ ms}^{-1}\) | M1, A1 | For using \(\frac{1}{2}mv^2 = mgh\) |
| (ii) \([WD = \frac{1}{2}x0.5(6^2 - 5^2)\) or \(0.5x10x1.8 = \frac{1}{2}x0.5x5^2]\) Work done is \(2.75 \text{ J}\) | M1, A1 | For using \(WD =\) loss of KE or \(KE_A + PE_A - WD = KE_C + PE_C\) |
(i) $[\frac{1}{2}v^2 = 10x \times 1.8]$ Speed is $6\text{ ms}^{-1}$ | M1, A1 | For using $\frac{1}{2}mv^2 = mgh$ | [2]
(ii) $[WD = \frac{1}{2}x0.5(6^2 - 5^2)$ or $0.5x10x1.8 = \frac{1}{2}x0.5x5^2]$ Work done is $2.75 \text{ J}$ | M1, A1 | For using $WD =$ loss of KE or $KE_A + PE_A - WD = KE_C + PE_C$ | [2]
---2\\
\includegraphics[max width=\textwidth, alt={}, center]{f0200d12-4ab0-4395-804c-e693f7f26507-2_301_1267_616_440}
The diagram shows the vertical cross-section $A B C$ of a fixed surface. $A B$ is a curve and $B C$ is a horizontal straight line. The part of the surface containing $A B$ is smooth and the part containing $B C$ is rough. $A$ is at a height of 1.8 m above $B C$. A particle of mass 0.5 kg is released from rest at $A$ and travels along the surface to $C$.\\
(i) Find the speed of the particle at $B$.\\
(ii) Given that the particle reaches $C$ with a speed of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find the work done against the resistance to motion as the particle moves from $B$ to $C$.
\hfill \mbox{\textit{CAIE M1 2010 Q2 [4]}}