| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Lighter particle on surface released, heavier hangs |
| Difficulty | Moderate -0.3 This is a standard M1 pulley system question with typical three-part structure: (a) find acceleration using F=ma for both particles (routine), (b) apply SUVAT (straightforward), (c) apply conservation of energy after string slackens (standard extension). While multi-step, it follows a well-practiced template with no novel insights required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| \(7g - T = 7a\), \(T - 3g = 3a\), \(4g = 10a\), \(a = \frac{4g}{10} = 3.92 \text{ m s}^{-2}\) (AG) | M1, M1, A1; dM1, A1 | M1: Three term equation of motion for one particle. Accept: \(7g - T = 7a\), \(3g - T = 3a\), \(T - 7g = 7a\), \(T - 3g = 3a\). M1: Three term equation of motion from the list above for the other particle. A1: Two consistent equations, that is: \(\frac{7g - T = 7a}{T - 3g = 3a}\) or \(\frac{T - 7g = 7a}{3g - T = 3a}\). dM1: Solving equations to find \(a\). A1: Obtaining 3.92 from consistent working. SC3: For whole string method. |
| Answer | Marks | Guidance |
|---|---|---|
| \(v^2 = 0^2 + 2 \times 3.92 \times 0.8 = 6.272\), \(v = 2.50 \text{ m s}^{-1}\) | M1A1; A1 | M1: Using \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = 0.8\) or 80 and \(a = 3.92\). A1: Correct equation. A1: Correct speed. Accept 2.5 or AWRT 2.50. |
| Answer | Marks | Guidance |
|---|---|---|
| \(0^2 = 6.272 + 2 \times (-9.8)s\), \(s = \frac{6.272}{19.6} = 0.32 \text{ m}\), Total height = \(32 + 80 = 112 \text{ cm}\) | M1A1; A1; B1 | M1: Using \(v^2 = u^2 + 2as\) with \(v = 0\), their value for \(v\) from (b) for \(u\) and \(a = \pm 9.8\). A1: Correct equation. Allow 6.25 for 2.5². A1: Obtaining AWRT ±0.32 from correct working. B1: Adding 80 or 0.8 to the height from their intermediate working. Must use same units and obtain an answer greater than 80 or 0.8 depending on units used. |
**5. (a)**
$7g - T = 7a$, $T - 3g = 3a$, $4g = 10a$, $a = \frac{4g}{10} = 3.92 \text{ m s}^{-2}$ (AG) | M1, M1, A1; dM1, A1 | M1: Three term equation of motion for one particle. Accept: $7g - T = 7a$, $3g - T = 3a$, $T - 7g = 7a$, $T - 3g = 3a$. M1: Three term equation of motion from the list above for the other particle. A1: Two consistent equations, that is: $\frac{7g - T = 7a}{T - 3g = 3a}$ or $\frac{T - 7g = 7a}{3g - T = 3a}$. dM1: Solving equations to find $a$. A1: Obtaining 3.92 from consistent working. SC3: For whole string method. | 5 marks
**5. (b)**
$v^2 = 0^2 + 2 \times 3.92 \times 0.8 = 6.272$, $v = 2.50 \text{ m s}^{-1}$ | M1A1; A1 | M1: Using $v^2 = u^2 + 2as$ with $u = 0$, $s = 0.8$ or 80 and $a = 3.92$. A1: Correct equation. A1: Correct speed. Accept 2.5 or AWRT 2.50. | 3 marks
**5. (c)**
$0^2 = 6.272 + 2 \times (-9.8)s$, $s = \frac{6.272}{19.6} = 0.32 \text{ m}$, Total height = $32 + 80 = 112 \text{ cm}$ | M1A1; A1; B1 | M1: Using $v^2 = u^2 + 2as$ with $v = 0$, their value for $v$ from (b) for $u$ and $a = \pm 9.8$. A1: Correct equation. Allow 6.25 for 2.5². A1: Obtaining AWRT ±0.32 from correct working. B1: Adding 80 or 0.8 to the height from their intermediate working. Must use same units and obtain an answer greater than 80 or 0.8 depending on units used. | 4 marks
**Total for Question 5: 12 marks**
---5 Two particles, of masses 3 kg and 7 kg , are connected by a light inextensible string that passes over a smooth peg. The 3 kg particle is held at ground level with the string above it taut and vertical. The 7 kg particle is at a height of 80 cm above ground level, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{5dd17095-18a6-470b-a24a-4456c8e3ed31-10_469_600_486_721}
The 3 kg particle is then released from rest.
\begin{enumerate}[label=(\alph*)]
\item By forming two equations of motion, show that the magnitude of the acceleration of the particles is $3.92 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\item Find the speed of the 7 kg particle just before it hits the ground.
\item When the 7 kg particle hits the ground, the string becomes slack and in the subsequent motion the 3 kg particle does not hit the peg.
Find the maximum height of the 3 kg particle above the ground.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2016 Q5 [4]}}