| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Difficulty | Standard +0.3 This is a straightforward application of the power-force-velocity relationship (P=Fv) combined with Newton's second law on an incline. Part (i) requires resolving forces and using F=ma to find driving force, then P=Fv. Part (ii) is direct substitution at constant speed (acceleration=0). Both parts are standard textbook exercises with clear methods and minimal problem-solving required, making it slightly easier than average. |
| Spec | 3.03c Newton's second law: F=ma one dimension6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| 3(i) | 1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 24000 × (0.2) | 1 | A1 |
| Power = Fv = 20561 × 25 | 1 | M1 |
| Power = 514 kW | 1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3(ii) | DF = 3200 + 24000g sin 3 | |
| [=15761] | 1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| v = 500000 / 15761 = 31.7 ms–1 | 1 | A1 |
Question 3:
--- 3(i) ---
3(i) | 1 | M1 | For applying Newton’s second law to the lorry up the
hill
F – 24000g sin 3 – 3200 =
24000 × (0.2) | 1 | A1 | [F = 20561]
Power = Fv = 20561 × 25 | 1 | M1 | Using P = Fv
Power = 514 kW | 1 | A1
4
--- 3(ii) ---
3(ii) | DF = 3200 + 24000g sin 3
[=15761] | 1 | M1 | Using Newton’s second law up the hill in the steady
case
v = 500000 / 15761 = 31.7 ms–1 | 1 | A1 | P = Fv so v = P / F
2A lorry of mass 24 000 kg is travelling up a hill which is inclined at 3° to the horizontal. The power developed by the lorry's engine is constant, and there is a constant resistance to motion of 3200 N.
\begin{enumerate}[label=(\roman*)]
\item When the speed of the lorry is 25 m s$^{-1}$, its acceleration is 0.2 m s$^{-2}$. Find the power developed by the lorry's engine. [4]
\item Find the steady speed at which the lorry moves up the hill if the power is 500 kW and the resistance remains 3200 N. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 Q3 [6]}}