| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find steady/maximum speed given power |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question testing standard power-force-velocity relationships (P=Fv) and Newton's second law on an incline. All three parts use direct formula application with no problem-solving insight required—slightly easier than average due to the routine nature and clear structure, though the multi-part format and incline component prevent it from being trivial. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Power \(= 1150 \times 12 = 13\,800\) W | B1 | For use of \(P = F \times v\). Allow 13.8 kW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Driving force \(= \dfrac{25000}{12}\) | B1 | Using \(F = \dfrac{P}{v}\) |
| \(\dfrac{25000}{12} - 1150 - 3700g\sin 4 = 3700a\) | M1 | For applying Newton's 2nd law up the slope, 4 terms |
| \(a = -0.445\) m s\(^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{25000}{v} - 1150 - 3700g\sin 4 = 0\) | M1 | For stating the equation for constant \(v\), with 3 terms, and solving for \(v\) |
| \(v = 6.70\) m s\(^{-1}\) | A1 |
## Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Power $= 1150 \times 12 = 13\,800$ W | B1 | For use of $P = F \times v$. Allow 13.8 kW |
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## Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Driving force $= \dfrac{25000}{12}$ | B1 | Using $F = \dfrac{P}{v}$ |
| $\dfrac{25000}{12} - 1150 - 3700g\sin 4 = 3700a$ | M1 | For applying Newton's 2nd law up the slope, 4 terms |
| $a = -0.445$ m s$^{-2}$ | A1 | |
---
## Question 2(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{25000}{v} - 1150 - 3700g\sin 4 = 0$ | M1 | For stating the equation for constant $v$, with 3 terms, and solving for $v$ |
| $v = 6.70$ m s$^{-1}$ | A1 | |
---2 A tractor of mass 3700 kg is travelling along a straight horizontal road at a constant speed of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The total resistance to motion is 1150 N .\\
(i) Find the power output of the tractor's engine.\\
The tractor comes to a hill inclined at $4 ^ { \circ }$ above the horizontal. The power output is increased to 25 kW and the resistance to motion is unchanged.\\
(ii) Find the deceleration of the tractor at the instant it begins to climb the hill.\\
(iii) Find the constant speed that the tractor could maintain on the hill when working at this power.\\
\hfill \mbox{\textit{CAIE M1 2017 Q2 [6]}}