| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on slope then horizontal |
| Difficulty | Standard +0.3 This is a straightforward energy conservation problem with two parts: (i) uses pure conservation of mechanical energy with no friction, and (ii) adds basic work done against friction. Both require standard application of formulas (mgh = ½mv² and friction work = μmgd) with simple arithmetic. The problem is slightly above average difficulty only because it has two parts and requires careful bookkeeping of energy changes. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(0.8g \times 4\) | B1 | |
| \([\frac{1}{2} \times 0.8v^2 = 32]\) | M1 | For using \(\frac{1}{2}mv_C^2 = \text{PE}_A\) and \(v_C = v_B\) |
| Speed at C is 8.94 ms⁻¹ | A1 | 3 |
| (ii) \([\text{Either } F = 0.3(0.8g) \text{ and } -2.4 = 0.8a \text{ or } F = 0.3(0.8g) \text{ and } WD = 2.4 \times 5]\) | M1 | For using \(F = \mu mg\) and either Newton's 2nd law to find a or \(WD = F \times BC\) |
| \([v^2 = \text{ans}(i)^2 - 2 \times 3 \times 5 \text{ or } \frac{1}{2} \times 0.8v^2 = 32 - 12]\) | M1 | For using either \(v^2 = u^2 + 2as\) or \(\frac{1}{2}mv^2 = \text{PE loss} - WD\) by F |
| Speed at C is 7.07 ms⁻¹ | A1 | 3 |
(i) $0.8g \times 4$ | B1 | | For finding PE at A
$[\frac{1}{2} \times 0.8v^2 = 32]$ | M1 | For using $\frac{1}{2}mv_C^2 = \text{PE}_A$ and $v_C = v_B$
Speed at C is 8.94 ms⁻¹ | A1 | 3
(ii) $[\text{Either } F = 0.3(0.8g) \text{ and } -2.4 = 0.8a \text{ or } F = 0.3(0.8g) \text{ and } WD = 2.4 \times 5]$ | M1 | For using $F = \mu mg$ and either Newton's 2nd law to find a or $WD = F \times BC$
$[v^2 = \text{ans}(i)^2 - 2 \times 3 \times 5 \text{ or } \frac{1}{2} \times 0.8v^2 = 32 - 12]$ | M1 | For using either $v^2 = u^2 + 2as$ or $\frac{1}{2}mv^2 = \text{PE loss} - WD$ by F
Speed at C is 7.07 ms⁻¹ | A1 | 34\\
\includegraphics[max width=\textwidth, alt={}, center]{155bc571-80e4-4c93-859f-bb150a109211-3_489_1041_258_552}\\
$A B C$ is a vertical cross-section of a surface. The part of the surface containing $A B$ is smooth and $A$ is 4 m higher than $B$. The part of the surface containing $B C$ is horizontal and the distance $B C$ is 5 m (see diagram). A particle of mass 0.8 kg is released from rest at $A$ and slides along $A B C$. Find the speed of the particle at $C$ in each of the following cases.\\
(i) The horizontal part of the surface is smooth.\\
(ii) The coefficient of friction between the particle and the horizontal part of the surface is 0.3 .
\hfill \mbox{\textit{CAIE M1 2011 Q4 [6]}}