| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Two particles over pulley, vertical strings |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem requiring Newton's second law for connected particles, solving simultaneous equations for acceleration and tension, then applying constant acceleration kinematics. While it involves multiple steps (4 marks for part c), each step follows routine procedures taught in M1 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(9g-T=9a\) | M1 | Equation for one particle |
| A1 | Correct equation | |
| \(T-5g=5a\) | M1 | Equation for other particle |
| A1 | Correct equation | |
| \(a=\frac{4g}{14}=2.8\text{ ms}^{-2}\) | A1 | Correct \(a\) from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(T-5g=5\times2.8\) | M1 | Substituting acceleration to find \(T\) |
| \(T=63\text{ N}\) | A1 | Correct tension |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(s=\frac{1}{2}\times2.8\times0.5^2\) | M1A1 | Constant acceleration equation with \(u=0\) and \(a\neq g\) to find \(s\); allow \(\pm\) answers |
| \(=0.35\text{ m}\) | A1 | Correct equation |
| \(\text{Total}=2\times0.35=0.7\text{ m}\) | A1ft | Correct distance; doubling their distance to get total distance apart |
## Question 4:
**Part (a)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $9g-T=9a$ | M1 | Equation for one particle |
| | A1 | Correct equation |
| $T-5g=5a$ | M1 | Equation for other particle |
| | A1 | Correct equation |
| $a=\frac{4g}{14}=2.8\text{ ms}^{-2}$ | A1 | Correct $a$ from correct working |
**Part (b)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $T-5g=5\times2.8$ | M1 | Substituting acceleration to find $T$ |
| $T=63\text{ N}$ | A1 | Correct tension |
**Part (c)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $s=\frac{1}{2}\times2.8\times0.5^2$ | M1A1 | Constant acceleration equation with $u=0$ and $a\neq g$ to find $s$; allow $\pm$ answers |
| $=0.35\text{ m}$ | A1 | Correct equation |
| $\text{Total}=2\times0.35=0.7\text{ m}$ | A1ft | Correct distance; doubling their distance to get total distance apart |
---4 Two particles, $A$ of mass 5 kg and $B$ of mass 9 kg , are attached to the ends of a light inextensible string. The string passes over a light smooth pulley as shown in the diagram. The particles are released from rest at the same height.\\
\includegraphics[max width=\textwidth, alt={}, center]{7e0585ea-062a-487c-8e39-37a4ed414ff8-3_378_287_1580_872}
\begin{enumerate}[label=(\alph*)]
\item By forming an equation of motion for each particle, show that the magnitude of the acceleration of each particle is $2.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\item Find the tension in the string.
\item When $B$ has been moving for 0.5 seconds it hits the floor. Find the height of $A$, above the floor, at this time. Assume that $A$ is still below the pulley when $B$ hits the floor.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2005 Q4 [11]}}