| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 13 |
| 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 work-energy-power question requiring straightforward application of P=Fv and F=ma. Part (a) involves resolving forces on an incline and calculating power (routine). Parts (b-c) apply the same principles on horizontal ground. Part (d) asks for a simple comment on modelling assumptions. All steps are textbook-standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{P}{v} - R - mg\sin\alpha = 0\) | M1 A1 | |
| \(\frac{P}{20} - 4400 - 40000(9.8) \cdot \frac{1}{20} = 0\) | M1 | |
| \(P = 20(4400 + 19600) = 480000\) W \(= 480\) kW | M1 A1 | |
| (b) \(\frac{P}{v} - R = ma \quad \therefore \frac{480000}{20} - 4400 = 40000a\) | M1 A1 | |
| \(a = 0.49\) ms\(^{-2}\) | A1 | |
| (c) at max. speed, \(a = 0\): \(\frac{P}{v} - R = 0\) | M1 | |
| \(\frac{480000}{v} - 4400 = 0\) so \(v = 109\) ms\(^{-1}\) (3sf) | M1 A1 | |
| (d) model not suitable – lorry unable to attain 109 ms\(^{-1}\) (≈ 245 mph) | B2 | (13) |
**(a)** $\frac{P}{v} - R - mg\sin\alpha = 0$ | M1 A1 |
$\frac{P}{20} - 4400 - 40000(9.8) \cdot \frac{1}{20} = 0$ | M1 |
$P = 20(4400 + 19600) = 480000$ W $= 480$ kW | M1 A1 |
**(b)** $\frac{P}{v} - R = ma \quad \therefore \frac{480000}{20} - 4400 = 40000a$ | M1 A1 |
$a = 0.49$ ms$^{-2}$ | A1 |
**(c)** at max. speed, $a = 0$: $\frac{P}{v} - R = 0$ | M1 |
$\frac{480000}{v} - 4400 = 0$ so $v = 109$ ms$^{-1}$ (3sf) | M1 A1 |
**(d)** model not suitable – lorry unable to attain 109 ms$^{-1}$ (≈ 245 mph) | B2 | (13)5. A lorry of mass 40 tonnes moves up a straight road inclined at an angle $\alpha$ to the horizontal where $\sin \alpha = \frac { 1 } { 20 }$. The lorry moves at a constant speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
In a model of the motion of the lorry, the non-gravitational resistance to motion is assumed to be constant and of magnitude 4400 N .
\begin{enumerate}[label=(\alph*)]
\item Show that the engine of the lorry is working at a rate of 480 kW .
The road becomes horizontal. The lorry's engine continues to work at the same rate and the resistance to motion is assumed to remain unchanged.
\item Find the initial acceleration of the lorry.
\item Find, correct to 3 significant figures, the maximum speed of the lorry.
\item Using your answer to part (c), comment on the suitability of the modelling assumption.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q5 [13]}}