| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on rough horizontal surface, particle hanging |
| Difficulty | Moderate -0.3 This is a standard M1 pulley system question with routine application of Newton's second law and kinematics. While it has multiple parts (5 total), each part follows directly from standard methods: friction force from F=μR, simultaneous equations for acceleration (given answer to verify), tension from either equation, and two straightforward SUVAT applications. The question requires no novel insight and is typical textbook fare, making it slightly easier than average for A-level. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03l Newton's third law: extend to situations requiring force resolution3.03o Advanced connected particles: and pulleys3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R = 14g\) | B1 | Normal reaction |
| \(F = \mu R = 0.25 \times 14g\) | M1 | Use of \(F = \mu R\) |
| \(F = 34.3 \text{ N}\) | A1 | CAO (accept 34 N) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For particle: \(6g - T = 6a\) | M1 A1 | Equation of motion for particle |
| For block: \(T - 34.3 = 14a\) | M1 A1 | Equation of motion for block |
| Adding: \(6g - 34.3 = 20a\) | DM1 | Solving simultaneously |
| \(a = \frac{58.8 - 34.3}{20} = \frac{24.5}{20} = 1.225 \approx 1.225 \text{ ms}^{-2}\) | A1 | Must be shown convincingly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T = 6g - 6(1.225)\) or \(T = 14(1.225) + 34.3\) | M1 | Substituting \(a\) back |
| \(T = 51.45 \text{ N}\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v^2 = u^2 + 2as = 0 + 2(1.225)(0.8)\) | M1 A1 | Using correct suvat with \(s=0.8\) |
| \(v = \sqrt{1.96} = 1.4 \text{ ms}^{-1}\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v^2 = 1.4^2 + 2(9.8)(0.5)\) | M1 A1 | Using \(u = 1.4\), \(s = 0.5\), \(g = 9.8\) |
| \(v = \sqrt{1.96 + 9.8} = \sqrt{11.76}\) | ||
| \(v = 3.43 \text{ ms}^{-1}\) | A1 | CAO |
# Question 5:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = 14g$ | B1 | Normal reaction |
| $F = \mu R = 0.25 \times 14g$ | M1 | Use of $F = \mu R$ |
| $F = 34.3 \text{ N}$ | A1 | CAO (accept 34 N) |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| For particle: $6g - T = 6a$ | M1 A1 | Equation of motion for particle |
| For block: $T - 34.3 = 14a$ | M1 A1 | Equation of motion for block |
| Adding: $6g - 34.3 = 20a$ | DM1 | Solving simultaneously |
| $a = \frac{58.8 - 34.3}{20} = \frac{24.5}{20} = 1.225 \approx 1.225 \text{ ms}^{-2}$ | A1 | Must be shown convincingly |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $T = 6g - 6(1.225)$ or $T = 14(1.225) + 34.3$ | M1 | Substituting $a$ back |
| $T = 51.45 \text{ N}$ | A1 | CAO |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $v^2 = u^2 + 2as = 0 + 2(1.225)(0.8)$ | M1 A1 | Using correct suvat with $s=0.8$ |
| $v = \sqrt{1.96} = 1.4 \text{ ms}^{-1}$ | A1 | CAO |
## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $v^2 = 1.4^2 + 2(9.8)(0.5)$ | M1 A1 | Using $u = 1.4$, $s = 0.5$, $g = 9.8$ |
| $v = \sqrt{1.96 + 9.8} = \sqrt{11.76}$ | | |
| $v = 3.43 \text{ ms}^{-1}$ | A1 | CAO |5 A block, of mass 14 kg , is held at rest on a rough horizontal surface. The coefficient of friction between the block and the surface is 0.25 . A light inextensible string, which passes over a fixed smooth peg, is attached to the block. The other end of the string is attached to a particle, of mass 6 kg , which is hanging at rest.\\
\includegraphics[max width=\textwidth, alt={}, center]{c022c936-72bc-4cf9-8f98-285f12c1d479-10_264_716_502_708}
The block is released and begins to accelerate.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the friction force acting on the block.
\item By forming two equations of motion, one for the block and one for the particle, show that the magnitude of the acceleration of the block and the particle is $1.225 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\item Find the tension in the string.
\item When the block is released, it is 0.8 metres from the peg. Find the speed of the block when it hits the peg.
\item When the block reaches the peg, the string breaks and the particle falls a further 0.5 metres to the ground. Find the speed of the particle when it hits the ground.\\
(3 marks)
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{c022c936-72bc-4cf9-8f98-285f12c1d479-11_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2009 Q5 [16]}}