| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | String breaks during motion |
| Difficulty | Standard +0.3 This is a standard M1 pulley system question requiring Newton's second law for connected particles, followed by kinematics after the string breaks. While multi-part with several steps (17 marks total), it follows a predictable template with no novel insights required—slightly easier than average due to its routine nature, but the extended calculation in part (c) prevents it from being significantly below average. |
| Spec | 3.02h Motion under gravity: vector form3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution |
| Answer | Marks |
|---|---|
| \((\downarrow)0.4g - T = 0.4a\) | M1 A1 |
| \((\uparrow)T - 0.3g = 0.3a\) | M1 A1 |
| solving for \(T\) | DM1 |
| \(T = 3.36\) or 3.4 or \(12g/35\) (N) | A1 |
| Answer | Marks |
|---|---|
| \(0.4g - 0.3g = 0.7a\) | DM1 |
| \(a = 1.4 \text{ m s}^{-2}\), \(g/7\) | A1 |
| Answer | Marks |
|---|---|
| \((\uparrow)v = u + at\) | M1 |
| \(v = 0.5 \times 1.4 = 0.7\) | A1 ft on a |
| \((\uparrow)s = ut + \frac{1}{2}at^2\) | M1 |
| \(s = 0.5 \times 1.4 \times 0.5^2 = 0.175\) | A1 ft on a |
| \((\downarrow)s = ut + \frac{1}{2}at^2\) | DM1 A1 ft |
| \(1.175 = -0.7t + 4.9t^2\) | |
| \(4.9t^2 - 0.7t - 1.175 = 0\) | |
| \(t = \frac{0.7 \pm \sqrt{0.7^2 + 19.6 \times 1.175}}{9.8}\) | DM1 A1 cao |
| \(= 0.5663...\text{ or } -...\) | |
| Ans 0.57 or 0.566 s | A1 cao |
## (a)
$(\downarrow)0.4g - T = 0.4a$ | M1 A1 |
$(\uparrow)T - 0.3g = 0.3a$ | M1 A1 |
solving for $T$ | DM1 |
$T = 3.36$ or 3.4 or $12g/35$ (N) | A1 |
**Total: (6)**
## (b)
$0.4g - 0.3g = 0.7a$ | DM1 |
$a = 1.4 \text{ m s}^{-2}$, $g/7$ | A1 |
**Total: (2)**
## (c)
$(\uparrow)v = u + at$ | M1 |
$v = 0.5 \times 1.4 = 0.7$ | A1 ft on a |
$(\uparrow)s = ut + \frac{1}{2}at^2$ | M1 |
$s = 0.5 \times 1.4 \times 0.5^2 = 0.175$ | A1 ft on a |
$(\downarrow)s = ut + \frac{1}{2}at^2$ | DM1 A1 ft |
$1.175 = -0.7t + 4.9t^2$ | |
$4.9t^2 - 0.7t - 1.175 = 0$ | |
$t = \frac{0.7 \pm \sqrt{0.7^2 + 19.6 \times 1.175}}{9.8}$ | DM1 A1 cao |
$= 0.5663...\text{ or } -...$ | |
Ans 0.57 or 0.566 s | A1 cao |
**Total: (9) [17]**\includegraphics{figure_3}
Two particles $A$ and $B$ have mass 0.4 kg and 0.3 kg respectively. The particles are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed above a horizontal floor. Both particles are held, with the string taut, at a height of 1 m above the floor, as shown in Figure 3. The particles are released from rest and in the subsequent motion $B$ does not reach the pulley.
\begin{enumerate}[label=(\alph*)]
\item Find the tension in the string immediately after the particles are released.
[6]
\item Find the acceleration of $A$ immediately after the particles are released.
[2]
\end{enumerate}
When the particles have been moving for 0.5 s, the string breaks.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the further time that elapses until $B$ hits the floor.
[9]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q8 [17]}}