| 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 | Up and down hill: two equations |
| Difficulty | Moderate -0.8 This is a straightforward mechanics problem requiring standard application of P=Fv and resolving forces on an incline. Part (i) uses constant speed (equilibrium), part (ii) adds F=ma. Both parts follow textbook procedures with no conceptual challenges, making it easier than average but not trivial due to the two-part structure and numerical computation involved. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([F = 1480 + 7850g\sin 3]\ (= 5588)\) | M1 | |
| \(\left[\frac{P}{10} = 1480 + 7850g\sin 3\right] \rightarrow P = \ldots\) | M1 | Using \(P = Fv\) and solving for \(P\) |
| Power \(= 55\,900\) W | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([F + 7850g\sin 3 - 1480 = 7850 \times 0.8]\ (F = 3652)\) | M1 | Use of Newton's Second Law |
| \(\left[\frac{P}{15} + 7850g\sin 3 - 1480 = 7850 \times 0.8\right] \rightarrow P = \ldots\) | M1 | Using \(P = Fv\) and solving for \(P\) |
| Power \(= 54800\) W | A1 | |
| 3 |
# Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[F = 1480 + 7850g\sin 3]\ (= 5588)$ | **M1** | |
| $\left[\frac{P}{10} = 1480 + 7850g\sin 3\right] \rightarrow P = \ldots$ | **M1** | Using $P = Fv$ and solving for $P$ |
| Power $= 55\,900$ W | **A1** | |
| | **3** | |
# Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[F + 7850g\sin 3 - 1480 = 7850 \times 0.8]\ (F = 3652)$ | **M1** | Use of Newton's Second Law |
| $\left[\frac{P}{15} + 7850g\sin 3 - 1480 = 7850 \times 0.8\right] \rightarrow P = \ldots$ | **M1** | Using $P = Fv$ and solving for $P$ |
| Power $= 54800$ W | **A1** | |
| | **3** | |
---2 A lorry of mass 7850 kg travels on a straight hill which is inclined at an angle of $3 ^ { \circ }$ to the horizontal. There is a constant resistance to motion of 1480 N .\\
(i) Find the power of the lorry's engine when the lorry is going up the hill at a constant speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Find the power of the lorry's engine at an instant when the lorry is going down the hill at a speed of $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ with an acceleration of $0.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
\hfill \mbox{\textit{CAIE M1 2017 Q2 [6]}}