| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Power from force and speed |
| Difficulty | Standard +0.3 This is a straightforward application of the work-energy principle with clearly stated values. Part (a) requires simple power calculation (P=Fv), while part (b) involves setting up an energy equation with given kinetic energy change, gravitational potential energy change, and work done by engine—all standard M1 techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P = 480 \times 24\), or e.g. \(\frac{P}{24} - 480 = 0\) | M1 | For \(\frac{P}{v} - F = 0\) or \(P = Fv\) oe. |
| \(P = 11.52\ \text{[kW]}\) | A1 | Allow 11.5 M1A0 for 11 520 or 11 500. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[KE_{before}\right] = \frac{1}{2}\times 1600\times 24^2\ [=460800]\); \(\left[KE_{after}\right] = \frac{1}{2}\times 1600\times 32^2\ [=819200]\) | B1 | For either correct. Do not allow \(\frac{1}{2}\times 1600\times(32-24)^2\). |
| \(\left[PE_{loss}\right] = 1600g\times 280\times 0.09\ [=1600g\times 25.2 = 403200]\) | B1 | Allow \(1600g\times 280\times\sin 5.16°\) or \(1600g\times 280\times\sin 5.2°\) but not simply \(1600g\times 280\times\sin\theta\) unless implied by correct final answer. |
| Total \(\text{WD} = 12000\times 10\ [=120000]\) | B1 | oe, e.g. \(12000 = \frac{\text{WD}}{10}\). |
| Work done against resistance \(= 280F =\) WD \(=\) W \(=\) oe: \(12000\times 10 + 1600g\times 280\times 0.09 - \frac{1}{2}\times 1600\times 32^2 + \frac{1}{2}\times 1200\times 24^2\) \([= 120000 + 403200 - 819200 + 460800]\) | M1 | Attempt at work-energy equation with 5 relevant terms (4 relevant terms plus work done against resistance); dimensionally correct. Allow sign errors. M0 for use of constant acceleration. Do not allow \(\frac{1}{2}\times 1600\times(32-24)^2\). |
| \(\text{WD} = 164\,800\ \text{[J]}\) | A1 | Or 164.8 kJ CAO but condone 165 kJ or 165 000 [J]. Not from use of constant acceleration or Newton's second law. ISW attempt to find force after correct WD found. |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P = 480 \times 24$, or e.g. $\frac{P}{24} - 480 = 0$ | M1 | For $\frac{P}{v} - F = 0$ or $P = Fv$ oe. |
| $P = 11.52\ \text{[kW]}$ | A1 | Allow 11.5 M1A0 for 11 520 or 11 500. |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[KE_{before}\right] = \frac{1}{2}\times 1600\times 24^2\ [=460800]$; $\left[KE_{after}\right] = \frac{1}{2}\times 1600\times 32^2\ [=819200]$ | B1 | For either correct. Do not allow $\frac{1}{2}\times 1600\times(32-24)^2$. |
| $\left[PE_{loss}\right] = 1600g\times 280\times 0.09\ [=1600g\times 25.2 = 403200]$ | B1 | Allow $1600g\times 280\times\sin 5.16°$ or $1600g\times 280\times\sin 5.2°$ but not simply $1600g\times 280\times\sin\theta$ unless implied by correct final answer. |
| Total $\text{WD} = 12000\times 10\ [=120000]$ | B1 | oe, e.g. $12000 = \frac{\text{WD}}{10}$. |
| Work done against resistance $= 280F =$ WD $=$ W $=$ oe: $12000\times 10 + 1600g\times 280\times 0.09 - \frac{1}{2}\times 1600\times 32^2 + \frac{1}{2}\times 1200\times 24^2$ $[= 120000 + 403200 - 819200 + 460800]$ | M1 | Attempt at work-energy equation with 5 relevant terms (4 relevant terms plus work done against resistance); dimensionally correct. Allow sign errors. M0 for use of constant acceleration. Do not allow $\frac{1}{2}\times 1600\times(32-24)^2$. |
| $\text{WD} = 164\,800\ \text{[J]}$ | A1 | Or 164.8 kJ CAO but condone 165 kJ or 165 000 [J]. Not from use of constant acceleration or Newton's second law. ISW attempt to find force after correct WD found. |4 A car has mass 1600 kg .
\begin{enumerate}[label=(\alph*)]
\item The car is moving along a straight horizontal road at a constant speed of $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is subject to a constant resistance of magnitude 480 N .
Find, in kW , the rate at which the engine of the car is working.\\
The car now moves down a hill inclined at an angle of $\theta$ to the horizontal, where $\sin \theta = 0.09$. The engine of the car is working at a constant rate of 12 kW . The speed of the car is $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the top of the hill. Ten seconds later the car has travelled 280 m down the hill and has speed $32 \mathrm {~ms} ^ { - 1 }$.
\item Given that the resistance is not constant, use an energy method to find the total work done against the resistance during the ten seconds.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q4 [7]}}