| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on smooth curved surface |
| Difficulty | Moderate -0.3 This is a standard energy conservation problem with two straightforward parts: (a) uses conservation of mechanical energy with a simple geometric setup (45° angle, halfway point), and (b) applies work-energy theorem with resistance. Both require routine application of formulas with minimal problem-solving insight, making it slightly easier than average for A-level mechanics. |
| Spec | 6.02a Work done: concept and definition6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| 5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or the |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a) | Attempt at finding PE lost | M1 |
| PE lost = 35g (4cos22.5 – 4cos45) | A1 |
| Answer | Marks |
|---|---|
| 2 | M1 |
| Speed = 4.16 ms–1 (4.1643...) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(b) | Use of the work-energy equation in the form: PE lost = KE gain + WD against resistance | M1 |
| Answer | Marks |
|---|---|
| 2 | A1 |
| X = 130 (130.05...) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or the
method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | Attempt at finding PE lost | M1
PE lost = 35g (4cos22.5 – 4cos45) | A1
1
× 35v2 = 35g (4cos22.5 – 4cos45)
2 | M1
Speed = 4.16 ms–1 (4.1643...) | A1
4
--- 5(b) ---
5(b) | Use of the work-energy equation in the form: PE lost = KE gain + WD against resistance | M1
1
× 35 × 42 = 35g (4 – 4cos45) – X
2 | A1
X = 130 (130.05...) | A1
3
Question | Answer | MarksA child of mass $35\text{ kg}$ is swinging on a rope. The child is modelled as a particle $P$ and the rope is modelled as a light inextensible string of length $4\text{ m}$. Initially $P$ is held at an angle of $45°$ to the vertical (see diagram).
\includegraphics{figure_5}
\begin{enumerate}[label=(\alph*)]
\item Given that there is no resistance force, find the speed of $P$ when it has travelled half way along the circular arc from its initial position to its lowest point. [4]
\item It is given instead that there is a resistance force. The work done against the resistance force as $P$ travels from its initial position to its lowest point is $X\text{ J}$. The speed of $P$ at its lowest point is $4\text{ m s}^{-1}$.
Find $X$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q5 [7]}}