| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Difficulty | Standard +0.3 This is a straightforward application of the work-energy principle in two parts. Part (i) requires calculating change in KE and adding work against resistance. Part (ii) involves setting up an energy equation with gravitational PE, resistance work, and driving force work. All values are given; students just need to apply standard formulas correctly with no conceptual challenges or novel problem-solving required. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Increase in KE \(= \frac{1}{2} \cdot 12500(25^2 - 17^2)\) | M1, A1 | For using \(KE = \frac{1}{2}mv^2\); Special case: max 1 mark if acceleration assumed constant: \(25^2 - 17^2 = 2ad\), \(F = 12500\times168/d\), \(KE\ \text{gain} = WD = Fd = 12500\times168\) → B1 |
| \([WD = 2100 + 5000]\) | M1 | For using \(WD\) by \(DF = KE\ \text{gain} + WD\ v\ \text{res}\) |
| Work done by driving force is 7100 kJ (or 7 100 000 J) | A1ft | ft only when units are consistent and both M marks are scored |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| M1 | For an equation with PE gain, WD by DF and WD v res (and KE loss if appropriate) in linear combination | |
| \(PE\ \text{gain} = (7100 + 3300) - (5000 + 4800\times500\div1000)\) or \(PE\ \text{gain} = 3300 + 2100 - 4800\times500\div1000\) | A1ft | Or equivalent in joules |
| \([3000\ 000 = 12500\times10h]\) | M1 | For solving \(mgh = \) gain PE found |
| Height is 24 m | A1 | Special case (constant acceleration): \(3300000/500 - 4800 - 12500\times10\sin\theta = 12500(-0.336)\) → B1; For using \(h = 500\sin\theta\) → M1; Height is 24 m → A1 |
## Question 5:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Increase in KE $= \frac{1}{2} \cdot 12500(25^2 - 17^2)$ | M1, A1 | For using $KE = \frac{1}{2}mv^2$; Special case: max 1 mark if acceleration assumed constant: $25^2 - 17^2 = 2ad$, $F = 12500\times168/d$, $KE\ \text{gain} = WD = Fd = 12500\times168$ → B1 |
| $[WD = 2100 + 5000]$ | M1 | For using $WD$ by $DF = KE\ \text{gain} + WD\ v\ \text{res}$ |
| Work done by driving force is 7100 kJ (or 7 100 000 J) | A1ft | ft only when units are consistent and both M marks are scored |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| | M1 | For an equation with PE gain, WD by DF and WD v res (and KE loss if appropriate) in linear combination |
| $PE\ \text{gain} = (7100 + 3300) - (5000 + 4800\times500\div1000)$ or $PE\ \text{gain} = 3300 + 2100 - 4800\times500\div1000$ | A1ft | Or equivalent in joules |
| $[3000\ 000 = 12500\times10h]$ | M1 | For solving $mgh = $ gain PE found |
| Height is 24 m | A1 | Special case (constant acceleration): $3300000/500 - 4800 - 12500\times10\sin\theta = 12500(-0.336)$ → B1; For using $h = 500\sin\theta$ → M1; Height is 24 m → A1 |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{f7a22c07-44e3-4891-be60-cbab772f45df-3_223_1456_1493_347}
A lorry of mass 12500 kg travels along a road that has a straight horizontal section $A B$ and a straight inclined section $B C$. The length of $B C$ is 500 m . The speeds of the lorry at $A , B$ and $C$ are $17 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $17 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively (see diagram).\\
(i) The work done against the resistance to motion of the lorry, as it travels from $A$ to $B$, is 5000 kJ . Find the work done by the driving force as the lorry travels from $A$ to $B$.\\
(ii) As the lorry travels from $B$ to $C$, the resistance to motion is 4800 N and the work done by the driving force is 3300 kJ . Find the height of $C$ above the level of $A B$.
\hfill \mbox{\textit{CAIE M1 2007 Q5 [8]}}