| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Two particles over pulley, vertical strings |
| Difficulty | Moderate -0.3 This is a standard M1 pulley problem requiring Newton's second law for connected particles, followed by straightforward kinematics. The acceleration is given to verify (not derive independently), and parts (d)(i)-(ii) use basic SUVAT equations. While it requires multiple steps, all techniques are routine textbook exercises with no novel problem-solving required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Peg is smooth | B1 | Correct assumption |
| Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| String is light | B1 | First correct assumption |
| String is inextensible or inelastic / Tension is the same throughout the string | B1 | Second correct assumption. Ignore any additional assumptions |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(11g - T = 11a\) | M1 | Equation of motion for \(A\), containing \(T\), \(11g\) or 107.8 and \(11a\) |
| A1 | Correct equation | |
| \(T - 9g = 9a\) | M1 | Equation of motion for \(B\) containing \(T\), \(9g\) or 88.2 and \(9a\) |
| A1 | Correct equation | |
| \(2g = 20a\), \(a = 0.98 \text{ ms}^{-2}\) AG | A1 | Correct acceleration from correct working. Do not penalise candidates who consistently use signs in the opposite direction, provided they give 0.98. If final answer is \(-0.98\) don't award final A1. Special Case: Whole String Method \(2g = 20a\) and \(a = 2g/20 = 0.98\) OE M1A1A1. Use of \(g = 9.81\) gives 0.981, first time award M1A1M1A1A0, don't penalise again |
| Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v = 0 + 0.98 \times 0.5 = 0.49 \text{ ms}^{-1}\) | M1 | Use of constant acceleration equation to find \(v\) with \(u = 0\), \(a = 0.98\) and \(t = 0.5\) |
| Correct \(v\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s = 0 + \frac{1}{2} \times 0.98 \times 0.5^2 = 0.1225 \text{ m}\) | M1 | Finding distance travelled with \(u = 0\), \(a = 0.98\) and \(t = 0.5\) |
| Correct distance | A1 | Accept 0.122 or 0.123 |
| OR \(0.49^2 = 0^2 + 2 \times 0.98s\), \(s = \frac{0.49^2}{2 \times 0.98} = 0.1225\) | (M1)(A1) | |
| \(d = 2 \times 0.1225 = 0.245 \text{ m}\) | M1 | Doubling distance or use of \(d/2\) in their original equation |
| Correct final distance; allow 0.244 or 0.246 | A1 | Use of \(0.5 \times 0.49 = 0.245\) scores zero unless justified |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Peg is smooth | B1 | Correct assumption |
| **Total: 1 mark** | | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| String is light | B1 | First correct assumption |
| String is inextensible or inelastic / Tension is the same throughout the string | B1 | Second correct assumption. Ignore any additional assumptions |
| **Total: 2 marks** | | |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $11g - T = 11a$ | M1 | Equation of motion for $A$, containing $T$, $11g$ or 107.8 and $11a$ |
| | A1 | Correct equation |
| $T - 9g = 9a$ | M1 | Equation of motion for $B$ containing $T$, $9g$ or 88.2 and $9a$ |
| | A1 | Correct equation |
| $2g = 20a$, $a = 0.98 \text{ ms}^{-2}$ **AG** | A1 | Correct acceleration from correct working. Do not penalise candidates who consistently use signs in the opposite direction, provided they give 0.98. If final answer is $-0.98$ don't award final A1. **Special Case:** Whole String Method $2g = 20a$ and $a = 2g/20 = 0.98$ OE M1A1A1. Use of $g = 9.81$ gives 0.981, first time award M1A1M1A1A0, don't penalise again |
| **Total: 5 marks** | | |
# Question 4:
## Part (d)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $v = 0 + 0.98 \times 0.5 = 0.49 \text{ ms}^{-1}$ | M1 | Use of constant acceleration equation to find $v$ with $u = 0$, $a = 0.98$ and $t = 0.5$ |
| Correct $v$ | A1 | |
## Part (d)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $s = 0 + \frac{1}{2} \times 0.98 \times 0.5^2 = 0.1225 \text{ m}$ | M1 | Finding distance travelled with $u = 0$, $a = 0.98$ and $t = 0.5$ |
| Correct distance | A1 | Accept 0.122 or 0.123 |
| **OR** $0.49^2 = 0^2 + 2 \times 0.98s$, $s = \frac{0.49^2}{2 \times 0.98} = 0.1225$ | (M1)(A1) | |
| $d = 2 \times 0.1225 = 0.245 \text{ m}$ | M1 | Doubling distance or use of $d/2$ in their original equation |
| Correct final distance; allow 0.244 or 0.246 | A1 | Use of $0.5 \times 0.49 = 0.245$ scores zero unless justified |
---4 Two particles, $A$ and $B$, are connected by a string that passes over a fixed peg, as shown in the diagram. The mass of $A$ is 9 kg and the mass of $B$ is 11 kg .\\
\begin{tikzpicture}[>=latex]
% Pulley
\draw[thick, fill=gray!20] (3,4.5) circle (0.4);
% Left string: from pulley down to mass A
\draw[thick] (2.6,4.5) -- (2.6,0);
% Right string: from pulley down to mass B
\draw[thick] (3.4,4.5) -- (3.4,2.5);
% Mass A (9 kg)
\fill (2.6,0) circle (5pt);
\node[right=8pt] at (2.6,0) {$A$\, (9\,kg)};
% Mass B (11 kg)
\fill (3.4,2.5) circle (5pt);
\node[right=8pt] at (3.4,2.5) {$B$\, (11\,kg)};
% Dashed reference lines
\draw[dashed] (0.5,2.5) -- (3.4,2.5);
\draw[dashed] (0.5,0) -- (2.6,0);
% Dimension arrow for d m
\draw[<->] (1,2.5) -- node[left] {$d$\,m} (1,0);
\end{tikzpicture}
The particles are released from rest in the position shown, where $B$ is $d$ metres higher than $A$. The motion of the particles is to be modelled using simple assumptions.
\begin{enumerate}[label=(\alph*)]
\item State one assumption that should be made about the peg.
\item State two assumptions that should be made about the string.
\item By forming an equation of motion for each of the particles $A$ and $B$, show that the acceleration of each particle has magnitude $0.98 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\item When the particles have been moving for 0.5 seconds, they are at the same level.
\begin{enumerate}[label=(\roman*)]
\item Find the speed of the particles at this time.
\item Find $d$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2009 Q4 [14]}}