| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find power at constant speed |
| Difficulty | Moderate -0.8 This is a straightforward application of the power-force-velocity relationship (P=Fv) with minimal problem-solving required. Part (i) is a direct 'show that' calculation, and part (ii) simply adds a component for the incline using basic trigonometry. The question involves standard M1 mechanics with clear signposting and routine methods throughout. |
| Spec | 3.03c Newton's second law: F=ma one dimension6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Resistance \(=\) Driving force \(= \dfrac{4080000}{85} = 48\,000\) N | B1 | Correct use of \(P = Fv\) and using \(\text{DF} = \text{Resistance}\) |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{DF} = \dfrac{P}{85}\) | B1 | \(\text{DF} = \dfrac{P}{v}\) |
| \(\text{DF} - 48\,000 - 490\,000\,g \times \dfrac{1}{200} = 0\) | M1 | For applying Newton's second law (3 terms) |
| \(P = 72\,500 \times 85 = 6.16\) MW | A1 | |
| Total: 3 |
## Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resistance $=$ Driving force $= \dfrac{4080000}{85} = 48\,000$ N | B1 | Correct use of $P = Fv$ and using $\text{DF} = \text{Resistance}$ |
| **Total: 1** | | |
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## Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{DF} = \dfrac{P}{85}$ | B1 | $\text{DF} = \dfrac{P}{v}$ |
| $\text{DF} - 48\,000 - 490\,000\,g \times \dfrac{1}{200} = 0$ | M1 | For applying Newton's second law (3 terms) |
| $P = 72\,500 \times 85 = 6.16$ MW | A1 | |
| **Total: 3** | | |2 A high-speed train of mass 490000 kg is moving along a straight horizontal track at a constant speed of $85 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The engines are supplying 4080 kW of power.\\
(i) Show that the resistance force is 48000 N .\\
(ii) The train comes to a hill inclined at an angle $\theta ^ { \circ }$ above the horizontal, where $\sin \theta ^ { \circ } = \frac { 1 } { 200 }$. Given that the resistance force is unchanged, find the power required for the train to keep moving at the same constant speed of $85 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
\hfill \mbox{\textit{CAIE M1 2018 Q2 [4]}}