| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle suspended by strings |
| Difficulty | Moderate -0.8 This is a straightforward equilibrium problem requiring basic resolution of forces in two directions and application of Newton's first law. Part (a) is guided, part (b) involves simple trigonometry with weight = 2g, and part (c) is a direct extension. All steps are standard M1 techniques with no problem-solving insight required. |
| Spec | 3.03b Newton's first law: equilibrium3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(T_1 \sin 35° = T_2 \sin 35°\) | M1 | Resolving two forces and forming an equation, with different tensions for each string |
| A1 | Correct result from correct working | |
| OR | ||
| \(T_1 \cos 55° = T_2 \cos 55°\) | ||
| \(T_1 = T_2\) | ||
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(T_1 \cos 35° + T_2 \cos 35° = 2 \times 9.8\) | M1 | Resolving forces to form a three term vertical equation |
| \(T_1 \cos 35° + T_1 \cos 35° = 2 \times 9.8\) | A1 | Correct equation |
| A1 | \(T_1\) or \(T_2\) eliminated correctly | |
| dM1 | Solving for \(T_1\) or \(T_2\) | |
| \(T_1 = \frac{2 \times 9.8}{2\cos 35°} = 12.0 \text{ N (to 3sf)}\) | A1 | Correct tension. Accept 12 N or 11.9 N |
| Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(2 \times 40 \cos 35° = 9.8m\) | M1 | Forming an equation with two tensions to find \(m\) |
| \(m = \frac{80 \cos 35°}{9.8} = 6.69 \text{ kg}\) | A1 | Correct equation |
| A1 | Correct mass. Accept 6.68 | |
| OR | ||
| \(m = \frac{40}{11.96} \times 2 = 6.69 \text{ kg}\) | (M1) (A1) (A1) | |
| Total: 10 marks |
**3(a)**
| $T_1 \sin 35° = T_2 \sin 35°$ | M1 | Resolving two forces and forming an equation, with different tensions for each string |
| | A1 | Correct result from correct working |
| **OR** | | |
| $T_1 \cos 55° = T_2 \cos 55°$ | | |
| $T_1 = T_2$ | | |
| **Total: 2 marks** | | |
**3(b)**
| $T_1 \cos 35° + T_2 \cos 35° = 2 \times 9.8$ | M1 | Resolving forces to form a three term vertical equation |
| $T_1 \cos 35° + T_1 \cos 35° = 2 \times 9.8$ | A1 | Correct equation |
| | A1 | $T_1$ or $T_2$ eliminated correctly |
| | dM1 | Solving for $T_1$ or $T_2$ |
| $T_1 = \frac{2 \times 9.8}{2\cos 35°} = 12.0 \text{ N (to 3sf)}$ | A1 | Correct tension. Accept 12 N or 11.9 N |
| **Total: 5 marks** | | |
**3(c)**
| $2 \times 40 \cos 35° = 9.8m$ | M1 | Forming an equation with two tensions to find $m$ |
| $m = \frac{80 \cos 35°}{9.8} = 6.69 \text{ kg}$ | A1 | Correct equation |
| | A1 | Correct mass. Accept 6.68 |
| **OR** | | |
| $m = \frac{40}{11.96} \times 2 = 6.69 \text{ kg}$ | (M1) (A1) (A1) | |
| **Total: 10 marks** | | |3 A sign, of mass 2 kg , is suspended from the ceiling of a supermarket by two light strings. It hangs in equilibrium with each string making an angle of $35 ^ { \circ }$ to the vertical, as shown in the diagram. Model the sign as a particle.\\
\includegraphics[max width=\textwidth, alt={}, center]{81f3753c-f148-44be-8b35-0a8e531016dd-2_424_385_1790_824}
\begin{enumerate}[label=(\alph*)]
\item By resolving forces horizontally, show that the tension is the same in each string.
\item Find the tension in each string.
\item If the tension in a string exceeds 40 N , the string will break. Find the mass of the heaviest sign that could be suspended as shown in the diagram.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2007 Q3 [10]}}