| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Two particles over pulley, vertical strings |
| Difficulty | Moderate -0.8 This is a standard M1 pulley question with straightforward application of Newton's second law. Part (a) requires simple equilibrium reasoning (T = mg), while part (b) involves setting up two equations of motion for connected particles—a routine textbook exercise with clear structure and no conceptual surprises. The calculation is guided ('show that T = 6.72N') rather than open-ended. |
| Spec | 3.03b Newton's first law: equilibrium3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(T = 0.6 \times 9.8 = 5.88\text{ N}\) or \(0.6g\) | B1 | Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Force \(= 2T = 11.76\text{ N}\) or \(11.8\text{ N}\) or \(1.2g\) | B1, B1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(Q: 0.8g - T = 0.8a\) | M1, A1 | Either equation |
| \(T - 0.6g = 0.6a\) | A1 | |
| \(0.2g = 1.4a\) | m1 | Alternative for m1 A1 if solving for \(T\) |
| \(a = 1.4\) | A1 | m1 method for solving, A1 accurate attempt |
| \(T = 6.72\text{ N}\) | A1 | Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Force \(= 2T = 13.44\text{ N}\) | B1 | Total: 1 |
## Question 7:
### Part (a)(i)
| Working | Marks | Guidance |
|---------|-------|----------|
| $T = 0.6 \times 9.8 = 5.88\text{ N}$ or $0.6g$ | B1 | Total: 1 | |
### Part (a)(ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| Force $= 2T = 11.76\text{ N}$ or $11.8\text{ N}$ or $1.2g$ | B1, B1 | Total: 2 | Magnitude; Direction |
### Part (b)(i)
| Working | Marks | Guidance |
|---------|-------|----------|
| $Q: 0.8g - T = 0.8a$ | M1, A1 | Either equation |
| $T - 0.6g = 0.6a$ | A1 | |
| $0.2g = 1.4a$ | m1 | Alternative for m1 A1 if solving for $T$ |
| $a = 1.4$ | A1 | m1 method for solving, A1 accurate attempt |
| $T = 6.72\text{ N}$ | A1 | Total: 6 | cao; SC whole string: to find $a$: $0.2g = 1.4a$ M1, $a = 1.4$ A1; to find $T$: M1 A1 |
### Part (b)(ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| Force $= 2T = 13.44\text{ N}$ | B1 | Total: 1 | cao |
---7 A builder ties two identical buckets, $P$ and $Q$, to the ends of a light inextensible rope. He hangs the rope over a smooth beam so that the buckets hang in equilibrium, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{c220e6c4-2676-4022-8301-7d720dc082b2-6_360_296_502_904}
The buckets are each of mass 0.6 kg .
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item State the magnitude of the tension in the rope.
\item State the magnitude and direction of the force exerted on the beam by the rope.
\end{enumerate}\item The bucket $Q$ is held at rest while a stone, of mass 0.2 kg , is placed inside it. The system is then released from rest and, in the subsequent motion, bucket $Q$ moves vertically downwards with the stone inside.
\begin{enumerate}[label=(\roman*)]
\item By forming an equation of motion for each bucket, show that the magnitude of the tension in the rope during the motion is 6.72 newtons, correct to three significant figures.
\item State the magnitude of the force exerted on the beam by the rope while the motion takes place.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2006 Q7 [5]}}