| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Towing system: horizontal road |
| Difficulty | Moderate -0.3 This is a straightforward mechanics problem requiring application of Newton's laws and the power equation P=Fv. Parts (a) and (b) are trivial (constant speed means zero tension and P=Rv). Part (c) requires setting up force equations for both objects and using P=Fv to find driving force, then solving simultaneously—standard A-level technique with no novel insight needed. The 5-mark allocation reflects routine multi-step working rather than conceptual difficulty. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.02l Power and velocity: P = Fv |
| Answer | Marks |
|---|---|
| 4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). |
| Answer | Marks | Guidance |
|---|---|---|
| 4(a) | Tension = 0 N | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(b) | Power = 0.22 =0.4W | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 4(c) | Driving force = 1.2/2 [= 0.6 N] | B1 |
| Use of Newton’s second law for locomotive or truck or system | M1 | Correct number of relevant terms. |
| Answer | Marks | Guidance |
|---|---|---|
| For system: DF – 0.2 =1 .2a | A1 | For any two correct. |
| For attempt to solve for T | M1 | From equations with correct number of relevant terms. |
| Answer | Marks | Guidance |
|---|---|---|
| 15 | A1 | Allow awrt 0.133 . |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | Tension = 0 N | B1 | May be implied.
1
--- 4(b) ---
4(b) | Power = 0.22 =0.4W | B1 | Use of power = Fv.
Allow without units.
1
Question | Answer | Marks | Guidance
--- 4(c) ---
4(c) | Driving force = 1.2/2 [= 0.6 N] | B1
Use of Newton’s second law for locomotive or truck or system | M1 | Correct number of relevant terms.
For locomotive: DF – 0.2 – T = 0.8a
For truck: T = 0.4a
For system: DF – 0.2 =1 .2a | A1 | For any two correct.
For attempt to solve for T | M1 | From equations with correct number of relevant terms.
Using their dimensionally correct DF.
1
May see a= .
3
2
T = N
15 | A1 | Allow awrt 0.133 .
5
Question | Answer | Marks | GuidanceA toy railway locomotive of mass 0.8 kg is towing a truck of mass 0.4 kg on a straight horizontal track at a constant speed of $2\,\text{m}\,\text{s}^{-1}$. There is a constant resistance force of magnitude 0.2 N on the locomotive, but no resistance force on the truck. There is a light rigid horizontal coupling connecting the locomotive and the truck.
\begin{enumerate}[label=(\alph*)]
\item State the tension in the coupling. [1]
\item Find the power produced by the locomotive's engine. [1]
The power produced by the locomotive's engine is now changed to 1.2 W.
\item Find the magnitude of the tension in the coupling at the instant that the locomotive begins to accelerate. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q4 [7]}}