| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Maximum speed on incline vs horizontal |
| Difficulty | Moderate -0.3 This is a standard M2 mechanics question requiring application of power = force × velocity and F = ma. Part (a) involves resolving forces on an incline and calculating power (routine but multi-step). Parts (b) and (c) are direct applications of familiar principles with no novel insight required. Slightly easier than average due to straightforward setup and standard techniques, though the multi-part nature and calculation complexity keep it close to average. |
| Spec | 3.03d Newton's second law: 2D vectors6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(F = 2000 + 4800g \cdot \frac{1}{20} = 4352\) N | M1 A1, A1 | |
| \(P = 12 \times 4652\) W \(\approx 52.2\) kW | M1 A1 ft | |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(4800a = 4352 - 2000\) | M1 A1 ft | |
| \(a = 0.49\) m s\(^{-2}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Max speed \(\frac{52224}{V} = 2000\) | M1 A1 | |
| \(V \approx 26.1\) ms\(^{-1}\) | A1 | (3) |
## Question 5:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| $F = 2000 + 4800g \cdot \frac{1}{20} = 4352$ N | M1 A1, A1 | |
| $P = 12 \times 4652$ W $\approx 52.2$ kW | M1 A1 ft | |
| | **(5)** | |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $4800a = 4352 - 2000$ | M1 A1 ft | |
| $a = 0.49$ m s$^{-2}$ | A1 | **(3)** |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| Max speed $\frac{52224}{V} = 2000$ | M1 A1 | |
| $V \approx 26.1$ ms$^{-1}$ | A1 | **(3)** |
---5. A straight road is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 20 }$. A lorry of mass 4800 kg moves up the road at a constant speed of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The non-gravitational resistance to the motion of the lorry is constant and has magnitude 2000 N .
\begin{enumerate}[label=(\alph*)]
\item Find, in kW to 3 significant figures, the rate of working of the lorry's engine.\\
(5)
The road becomes horizontal. The lorry's engine continues to work at the same rate and the resistance to motion remains the same.
Find
\item the acceleration of the lorry immediately after the road becomes horizontal,\\
(3)
\item the maximum speed, in $\mathrm { m } \mathrm { s } ^ { - 1 }$ to 3 significant figures, at which the lorry will go along the horizontal road.\\
(3)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q5 [11]}}