| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | November |
| Marks | 5 |
| 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 power-force-velocity question requiring direct application of P=Fv and F=ma. Part (a) uses constant speed (so driving force equals resistance) and part (b) requires finding net force from power then applying Newton's second law. Both are standard M1 exercises with no conceptual challenges beyond routine formula manipulation. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| 2(a) | P = 350 × 20 | M1 |
| P = 7 kW | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2(b) | 15 000 = DF × 20 [DF = 750] | B1 |
| DF – 350 = 1400a | M1 | Use Newton’s 2nd law, 3 terms |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | A1 | a = 0.286 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 2:
--- 2(a) ---
2(a) | P = 350 × 20 | M1 | Using P = Fv
P = 7 kW | A1
2
--- 2(b) ---
2(b) | 15 000 = DF × 20 [DF = 750] | B1 | Using P = Fv
DF – 350 = 1400a | M1 | Use Newton’s 2nd law, 3 terms
2
a = ms–2
7 | A1 | a = 0.286
3
Question | Answer | Marks | GuidanceA car of mass 1400 kg is moving along a straight horizontal road against a resistance of magnitude 350 N.
\begin{enumerate}[label=(\alph*)]
\item Find, in kW, the rate at which the engine of the car is working when it is travelling at a constant speed of $20 \text{ m s}^{-1}$. [2]
\item Find the acceleration of the car when its speed is $20 \text{ m s}^{-1}$ and the engine is working at 15 kW. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q2 [5]}}