| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | String breaks during motion |
| Difficulty | Moderate -0.3 This is a standard M1 pulley question with straightforward application of Newton's second law and SUVAT equations. Part (a) is a routine 'show that' for acceleration, part (b) is direct substitution for tension, and part (c) involves standard kinematics after string breaks. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03l Newton's third law: extend to situations requiring force resolution3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| For particle \(A\) (moving down): \(12g - T = 12a\) | M1 | Equation of motion for either particle |
| For particle \(B\) (moving up): \(T - 8g = 8a\) | A1 | Correct equation for both |
| Adding: \(4g = 20a\) | M1 | Eliminating \(T\) |
| \(a = \frac{4g}{20} = \frac{g}{5} = \frac{9.8}{5} = 1.96 \text{ ms}^{-2}\) | A1 | Correct value shown |
| Both equations correct | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(T = 8(g + a) = 8(9.8 + 1.96)\) | M1 | Substituting back |
| \(T = 94.08 \approx 94.1 \text{ N}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(v = u + at = 0 + 1.96 \times 2\) | M1 | Using \(v = u + at\) with \(t=2\) |
| \(v = 3.92 \text{ ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| After string breaks, \(A\) moves upward with initial velocity \(3.92\) ms\(^{-1}\), decelerating at \(g\) | M1 | Correct identification of motion |
| \(v^2 = 3.92^2 - 2(9.8)(4)\) | M1 | Using \(v^2 = u^2 + 2as\) with \(s=4\) upward |
| \(v^2 = 15.3664 - 78.4 < 0\), so \(A\) must use \(v^2 = u^2 + 2as\) falling from max height | DM1 | |
| Max height above break point: \(h = \frac{3.92^2}{2 \times 9.8} = 0.784\) m, total height \(= 4.784\) m | A1 | |
| \(v^2 = 2(9.8)(4.784)\), \(v = 9.68 \text{ ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| At string break, \(B\) has velocity \(3.92\) ms\(^{-1}\) upward, height \(4\) m | M1 | Correct initial conditions for \(B\) |
| After break, \(B\) is in free fall (only gravity): \(-4 = 3.92t - \frac{1}{2}(9.8)t^2\) | M1 | Using \(s = ut + \frac{1}{2}at^2\), \(s = -4\) |
| \(4.9t^2 - 3.92t - 4 = 0\) | M1 | Forming correct equation |
| \(t = \frac{3.92 \pm \sqrt{15.3664 + 78.4}}{9.8}\) | DM1 | Solving quadratic |
| \(t = \frac{3.92 + 9.68}{9.8} = \frac{13.6}{9.8} \approx 1.39 \text{ s}\) | A1 | Taking positive root |
# Question 6:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| For particle $A$ (moving down): $12g - T = 12a$ | M1 | Equation of motion for either particle |
| For particle $B$ (moving up): $T - 8g = 8a$ | A1 | Correct equation for both |
| Adding: $4g = 20a$ | M1 | Eliminating $T$ |
| $a = \frac{4g}{20} = \frac{g}{5} = \frac{9.8}{5} = 1.96 \text{ ms}^{-2}$ | A1 | Correct value shown |
| Both equations correct | B1 | |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $T = 8(g + a) = 8(9.8 + 1.96)$ | M1 | Substituting back |
| $T = 94.08 \approx 94.1 \text{ N}$ | A1 | |
## Part (c)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $v = u + at = 0 + 1.96 \times 2$ | M1 | Using $v = u + at$ with $t=2$ |
| $v = 3.92 \text{ ms}^{-1}$ | A1 | |
## Part (c)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| After string breaks, $A$ moves upward with initial velocity $3.92$ ms$^{-1}$, decelerating at $g$ | M1 | Correct identification of motion |
| $v^2 = 3.92^2 - 2(9.8)(4)$ | M1 | Using $v^2 = u^2 + 2as$ with $s=4$ upward |
| $v^2 = 15.3664 - 78.4 < 0$, so $A$ must use $v^2 = u^2 + 2as$ falling from max height | DM1 | |
| Max height above break point: $h = \frac{3.92^2}{2 \times 9.8} = 0.784$ m, total height $= 4.784$ m | A1 | |
| $v^2 = 2(9.8)(4.784)$, $v = 9.68 \text{ ms}^{-1}$ | A1 | |
## Part (c)(iii):
| Working | Mark | Guidance |
|---------|------|----------|
| At string break, $B$ has velocity $3.92$ ms$^{-1}$ upward, height $4$ m | M1 | Correct initial conditions for $B$ |
| After break, $B$ is in free fall (only gravity): $-4 = 3.92t - \frac{1}{2}(9.8)t^2$ | M1 | Using $s = ut + \frac{1}{2}at^2$, $s = -4$ |
| $4.9t^2 - 3.92t - 4 = 0$ | M1 | Forming correct equation |
| $t = \frac{3.92 \pm \sqrt{15.3664 + 78.4}}{9.8}$ | DM1 | Solving quadratic |
| $t = \frac{3.92 + 9.68}{9.8} = \frac{13.6}{9.8} \approx 1.39 \text{ s}$ | A1 | Taking positive root |
---6 Two particles, $A$ and $B$, have masses 12 kg and 8 kg respectively. They are connected by a light inextensible string that passes over a smooth fixed peg, as shown in the diagram.
$$A ( 12 \mathrm {~kg} )$$
The particles are released from rest and move vertically. Assume that there is no air resistance.
\begin{enumerate}[label=(\alph*)]
\item By forming two equations of motion, show that the magnitude of the acceleration of each particle is $1.96 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\item Find the tension in the string.
\item After the particles have been moving for 2 seconds, both particles are at a height of 4 metres above a horizontal surface. When the particles are in this position, the string breaks.
\begin{enumerate}[label=(\roman*)]
\item Find the speed of particle $A$ when the string breaks.
\item Find the speed of particle $A$ when it hits the surface.
\item Find the time that it takes for particle $B$ to reach the surface after the string breaks. Assume that particle $B$ does not hit the peg.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-13_2484_1709_223_153}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2010 Q6 [17]}}