| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done against friction/resistance - inclined plane or slope |
| Difficulty | Moderate -0.8 This is a straightforward application of conservation of energy and work-energy principles. Part (i) uses basic energy conservation (KE + PE), while part (ii) applies the work-energy theorem with given values. Both parts require only direct substitution into standard formulas with no problem-solving insight or complex multi-step reasoning. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| [\(\frac{1}{2}mv^2 - \frac{1}{2}m7^2 = \text{mgxs}\)] | M1 | |
| M1 | For equation from KE gain = PE loss (3 terms) | |
| Speed is 12.2ms⁻¹ | A1 | 3 |
| SR for candidates who treat AB as straight and vertical (max 1mark out of 3) | ||
| \(v^2 = 7^2 + 2gs \Rightarrow v = 12.2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | For using WD = PE loss − KE gain or WD = KE at B in (i) − actual KE at B | |
| WD = 0.35x10x5 − ½ 0.35(11² − 7²) or WD = ½ 0.35(12.2² − 11²) | A1ft | ft wrong v in part (i) or for 12.2 scored by B1in (i) |
| Work done is 4.9 J | A1 | 3 |
| SR for candidates who treat AB as straight and vertical, and resistance as constant (max 1mark out of 3) | ||
| \(a = 7.2 \text{ ms}^{-2}, R = 0.98 \text{ N, WD} = 4.9 \text{ J}\) | B1 | |
| SR for candidates who write 'Resistance =' instead of 'WD =' (max 2/3) | ||
| 0.35x10x5 − ½ 0.35(11² − 7²) or ½ 0.35(12.2² − 11²) seen | B1 | |
| Answer 4.9J (NB J seen) | B1 |
**(i)**
[$\frac{1}{2}mv^2 - \frac{1}{2}m7^2 = \text{mgxs}$] | M1 |
| M1 | For equation from KE gain = PE loss (3 terms)
Speed is 12.2ms⁻¹ | A1 | 3
| | | SR for candidates who treat AB as straight and vertical (max 1mark out of 3)
| | | $v^2 = 7^2 + 2gs \Rightarrow v = 12.2$ | B1 |
**(ii)**
| M1 | For using WD = PE loss − KE gain or WD = KE at B in (i) − actual KE at B
WD = 0.35x10x5 − ½ 0.35(11² − 7²) or WD = ½ 0.35(12.2² − 11²) | A1ft | ft wrong v in part (i) or for 12.2 scored by B1in (i)
Work done is 4.9 J | A1 | 3 | This mark is not available if v = 12.2 is used, having been scored by B1 in part (i)
| | | SR for candidates who treat AB as straight and vertical, and resistance as constant (max 1mark out of 3)
| | | $a = 7.2 \text{ ms}^{-2}, R = 0.98 \text{ N, WD} = 4.9 \text{ J}$ | B1 |
| | | SR for candidates who write 'Resistance =' instead of 'WD =' (max 2/3)
| | | 0.35x10x5 − ½ 0.35(11² − 7²) or ½ 0.35(12.2² − 11²) seen | B1 |
| | | Answer 4.9J (NB J seen) | B1 |\includegraphics{figure_4}
The diagram shows the vertical cross-section of a surface. $A$ and $B$ are two points on the cross-section, and $A$ is 5 m higher than $B$. A particle of mass $0.35$ kg passes through $A$ with speed $7 \text{ m s}^{-1}$, moving on the surface towards $B$.
\begin{enumerate}[label=(\roman*)]
\item Assuming that there is no resistance to motion, find the speed with which the particle reaches $B$. [3]
\item Assuming instead that there is a resistance to motion, and that the particle reaches $B$ with speed $11 \text{ m s}^{-1}$, find the work done against this resistance as the particle moves from $A$ to $B$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2007 Q4 [6]}}