| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Constant speed up/down incline |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring standard application of power equations (P=Fv) and work-energy principles. Part (a) involves resolving forces at constant speed on an incline, while part (b) applies the work-energy theorem with given values. The calculations are direct with no conceptual tricks, making it slightly easier than average. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Driving force \(= DF = \dfrac{960\,000}{30}\) | B1 | Allow for \(960\,000 = DF \times 30\) |
| \(DF - 75000g \times \sin\alpha - R = 0\) | M1 | Resolve forces along the slope. Must use a value for either \(\sin\alpha\) or \(\alpha\). |
| Resistance force \(= R = 24\,500\) N | A1 | Allow correct work with 24500 to 3 sf. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| WD by engine in \(60\ \text{s} = 900\,000 \times 60\ [= 54000000]\) | B1 | |
| \(KE_{init} = \frac{1}{2}\times 75000\times 30^2 \quad KE_{final} = \frac{1}{2}\times 75000\times v^2\) | B1 | For either correct expression for KE. |
| \(900000\times 60 + \frac{1}{2}\times 75000\times 30^2 = 46500000 + \frac{1}{2}\times 75000\times v^2\) | M1 | For use of the work-energy equation with 4 terms, correct dimensions. |
| Speed of engine after \(60\ \text{s} = v = 33.2\) ms\(^{-1}\) | A1 | Allow \(v = \sqrt{1100} = 10\sqrt{11}\) |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Driving force $= DF = \dfrac{960\,000}{30}$ | B1 | Allow for $960\,000 = DF \times 30$ |
| $DF - 75000g \times \sin\alpha - R = 0$ | M1 | Resolve forces along the slope. Must use a value for either $\sin\alpha$ or $\alpha$. |
| Resistance force $= R = 24\,500$ N | A1 | Allow correct work with 24500 to 3 sf. |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| WD by engine in $60\ \text{s} = 900\,000 \times 60\ [= 54000000]$ | B1 | |
| $KE_{init} = \frac{1}{2}\times 75000\times 30^2 \quad KE_{final} = \frac{1}{2}\times 75000\times v^2$ | B1 | For either correct expression for KE. |
| $900000\times 60 + \frac{1}{2}\times 75000\times 30^2 = 46500000 + \frac{1}{2}\times 75000\times v^2$ | M1 | For use of the work-energy equation with 4 terms, correct dimensions. |
| Speed of engine after $60\ \text{s} = v = 33.2$ ms$^{-1}$ | A1 | Allow $v = \sqrt{1100} = 10\sqrt{11}$ |
---5 A railway engine of mass 75000 kg is moving up a straight hill inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = 0.01$. The engine is travelling at a constant speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The engine is working at 960 kW . There is a constant force resisting the motion of the engine.
\begin{enumerate}[label=(\alph*)]
\item Find the resistance force.\\
The engine comes to a section of track which is horizontal. At the start of the section the engine is travelling at $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the power of the engine is now reduced to 900 kW . The resistance to motion is no longer constant, but in the next 60 s the work done against the resistance force is 46500 kJ .
\item Find the speed of the engine at the end of the 60 s .
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q5 [7]}}