| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on smooth curved surface |
| Difficulty | Moderate -0.8 This is a straightforward application of conservation of energy and work-energy principles with standard values. Part (i) uses PE = KE directly, part (ii) adds a simple work-done-against-resistance calculation. Both are routine M1 mechanics exercises requiring only direct substitution into familiar formulas with no problem-solving insight needed. |
| Spec | 6.02c Work by variable force: using integration6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| KE (gain) = \(\frac{1}{2} \times 0.15 \times 8^2\) | B1 | |
| For using PE loss = KE gain | M1 | |
| Height is 3.2 m | A1 | 3 marks |
| For using WD is difference in PE loss and KE gain | M1 | |
| \(WD = 0.15 \times 10 \times 4 - \frac{1}{2} \times 0.15 \times 6^2\) | A1 | |
| Work Done is 3.3 J | A1 | 3 marks |
| SR: For candidates who treat AB as if it is straight and vertical (implicitly or otherwise) | Max 2 out of 6 marks | |
| (i) \(a = 8^2 \div (2 \times 10) = 3.2\) | B1 | |
| (ii) \(a = 6^2 + (2 \times 4) = 4.5\) and \(R = 0.15 \times 10 - 0.15 \times 4.5 = 0.825\) and \(WD = 4 \times 0.825 = 3.3\) | B1 | 2 marks |
KE (gain) = $\frac{1}{2} \times 0.15 \times 8^2$ | B1 |
For using PE loss = KE gain | M1 |
Height is 3.2 m | A1 | 3 marks
For using WD is difference in PE loss and KE gain | M1 |
$WD = 0.15 \times 10 \times 4 - \frac{1}{2} \times 0.15 \times 6^2$ | A1 |
Work Done is 3.3 J | A1 | 3 marks
**SR:** For candidates who treat AB as if it is straight and vertical (implicitly or otherwise) | Max 2 out of 6 marks
(i) $a = 8^2 \div (2 \times 10) = 3.2$ | B1
(ii) $a = 6^2 + (2 \times 4) = 4.5$ and $R = 0.15 \times 10 - 0.15 \times 4.5 = 0.825$ and $WD = 4 \times 0.825 = 3.3$ | B1 | 2 marks
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{5cba3e17-3979-4c22-a415-2cdd60f09289-2_227_586_1631_781}
The diagram shows a vertical cross-section of a surface. $A$ and $B$ are two points on the cross-section. A particle of mass 0.15 kg is released from rest at $A$.\\
(i) Assuming that the particle reaches $B$ with a speed of $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and that there are no resistances to motion, find the height of $A$ above $B$.\\
(ii) Assuming instead that the particle reaches $B$ with a speed of $6 \mathrm {~ms} ^ { - 1 }$ and that the height of $A$ above $B$ is 4 m , find the work done against the resistances to motion.
\hfill \mbox{\textit{CAIE M1 2003 Q4 [6]}}