| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Car towing trailer, horizontal |
| Difficulty | Moderate -0.3 This is a standard M1 connected particles question requiring straightforward application of Newton's second law to two bodies. Parts (a) and (b) involve routine F=ma calculations with given values, while part (c) requires basic conceptual understanding of force components. Slightly easier than average due to clear setup and standard method. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03m Equilibrium: sum of resolved forces = 03.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - 800 = 1200 \times 0.4\) | M1 | Three term equation of motion for the car |
| \(T = 800 + 480 = 1280 \text{ N}\) | A1 | Correct equation |
| A1 | Correct tension. Treat calculation of two tensions as two methods unless one selected. Treat sum or difference of two tensions as an incorrect method | |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(3000 - 800 - F = 4000 \times 0.4\) | M1 | Four term equation of motion (truck or both) |
| \(F = 3000 - 800 - 1600 = 600 \text{ N}\) | A1 | Correct terms |
| A1 | Correct signs | |
| A1 | AG Correct resistance force from correct working | |
| OR | ||
| \(3000 - 1280 - F = 2800 \times 0.4\) | ||
| \(F = 3000 - 1280 - 1120 = 600 \text{ N}\) | ||
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Increase, because a greater tension would be needed so that the horizontal component would be the same as the tension above. | B1 | Greater |
| B1 | Reason. Second B1 dependent on the first B1 mark | |
| Total: 9 marks |
**4(a)**
| $T - 800 = 1200 \times 0.4$ | M1 | Three term equation of motion for the car |
| $T = 800 + 480 = 1280 \text{ N}$ | A1 | Correct equation |
| | A1 | Correct tension. Treat calculation of two tensions as two methods unless one selected. Treat sum or difference of two tensions as an incorrect method |
| **Total: 3 marks** | | |
**4(b)**
| $3000 - 800 - F = 4000 \times 0.4$ | M1 | Four term equation of motion (truck or both) |
| $F = 3000 - 800 - 1600 = 600 \text{ N}$ | A1 | Correct terms |
| | A1 | Correct signs |
| | A1 | AG Correct resistance force from correct working |
| **OR** | | |
| $3000 - 1280 - F = 2800 \times 0.4$ | | |
| $F = 3000 - 1280 - 1120 = 600 \text{ N}$ | | |
| **Total: 4 marks** | | |
**4(c)**
| Increase, because a greater tension would be needed so that the horizontal component would be the same as the tension above. | B1 | Greater |
| | B1 | Reason. Second B1 dependent on the first B1 mark |
| **Total: 9 marks** | | |4 A car, of mass 1200 kg , is connected by a tow rope to a truck, of mass 2800 kg . The truck tows the car in a straight line along a horizontal road. Assume that the tow rope is horizontal. A horizontal driving force of magnitude 3000 N acts on the truck. A horizontal resistance force of magnitude 800 N acts on the car. The car and truck accelerate at $0.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{81f3753c-f148-44be-8b35-0a8e531016dd-3_177_1002_580_513}
\begin{enumerate}[label=(\alph*)]
\item Find the tension in the tow rope.
\item Show that the magnitude of the horizontal resistance force acting on the truck is 600 N .
\item In fact, the tow rope is not horizontal. Assume that the resistance forces and the driving force are unchanged.
Is the tension in the tow rope greater or less than in part (a)?
Explain why.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2007 Q4 [9]}}