| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find acceleration on incline given power |
| Difficulty | Moderate -0.3 This is a straightforward application of the power equation P=Fv with Newton's second law on an incline. Part (i) requires resolving forces to find driving force, then calculating power. Part (ii) reverses this at constant speed (zero acceleration). Both parts use standard mechanics formulas with no conceptual challenges, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For applying Newton's second law to the lorry up the hill | |
| \(F - 24000g\sin 3 - 3200 = 24000 \times (0.2)\) | A1 | \([F = 20561]\) |
| Power \(= Fv = 20561 \times 25\) | M1 | Using \(P = Fv\) |
| Power \(= 514\) kW | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(DF = 3200 + 24000g\sin 3\ [=15761]\) | M1 | Using Newton's second law up the hill in the steady case |
| \(v = 500000/15761 = 31.7\ \text{ms}^{-1}\) | A1 (2) | \(P = Fv\) so \(v = P/F\) |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For applying Newton's second law to the lorry up the hill |
| $F - 24000g\sin 3 - 3200 = 24000 \times (0.2)$ | A1 | $[F = 20561]$ |
| Power $= Fv = 20561 \times 25$ | M1 | Using $P = Fv$ |
| Power $= 514$ kW | A1 (4) | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $DF = 3200 + 24000g\sin 3\ [=15761]$ | M1 | Using Newton's second law up the hill in the steady case |
| $v = 500000/15761 = 31.7\ \text{ms}^{-1}$ | A1 (2) | $P = Fv$ so $v = P/F$ |
---3 A lorry of mass 24000 kg is travelling up a hill which is inclined at $3 ^ { \circ }$ to the horizontal. The power developed by the lorry's engine is constant, and there is a constant resistance to motion of 3200 N .\\
(i) When the speed of the lorry is $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, its acceleration is $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Find the power developed by the lorry's engine.\\
(ii) Find the steady speed at which the lorry moves up the hill if the power is 500 kW and the resistance remains 3200 N .
\hfill \mbox{\textit{CAIE M1 2015 Q3 [6]}}