| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Three or more connected particles |
| Difficulty | Standard +0.3 This is a standard M1 pulley/connected particles question requiring systematic application of Newton's second law and kinematics. While it has multiple parts and involves two phases of motion, each step follows routine mechanics procedures: finding acceleration from SUVAT, applying F=ma with friction to both particles, and analyzing motion after force removal. The 'inextensible string' part (d) tests understanding rather than calculation. More straightforward than average A-level questions due to its predictable structure and standard techniques. |
| Spec | 3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(s = ut + \frac{1}{2}at^2 \Rightarrow 6 = \frac{1}{2}a \times 9\) giving \(a = 1\frac{1}{3} \text{ (ms}^{-2})\) | M1 A1 | (2) |
| (b) N2L for system: \(30 - \mu5g = 5a\) with f.t. their \(a\), giving \(\mu = \frac{14}{3g} = \frac{10}{21}\) or awrt 0.48 | M1 A1 f.t. DM1 A1 | (4) |
| (c) N2L for \(P\): \(T - \mu 2g = 2a\) with f.t. their \(\mu\), their \(a\), accepting symbols, giving \(T - \frac{14}{3g} \times 2g = 2 \times \frac{4}{3}\) leading to \(T = 12 \text{ (N)}\) | M1 A1 f.t. DM1 A1 | (4) |
| Alternatively N2L for \(Q\): \(30 - T - \mu3g = 3a\) leading to \(T = 12 \text{ (N)}\) | M1 A1 DM1 A1 | awrt 12 |
| (d) The acceleration of \(P\) and \(Q\) (or the whole of the system) is the same. | B1 | (1) |
| (e) \(v = u + at \Rightarrow v = \frac{4}{3} \times 3 = 4\) with N2L (for system or either particle) \(-5\mu g = 5a\) giving \(a = -\mu g\) and \(v = u + at \Rightarrow 0 = 4 - \mu gt\) leading to \(t = \frac{6}{7} \text{ (s)}\) | B1 f.t. on \(a\) M1 DM1 A1 | (4) [15] Accept 0.86, 0.857 |
**(a)** $s = ut + \frac{1}{2}at^2 \Rightarrow 6 = \frac{1}{2}a \times 9$ giving $a = 1\frac{1}{3} \text{ (ms}^{-2})$ | M1 A1 | (2)
**(b)** N2L for system: $30 - \mu5g = 5a$ with f.t. their $a$, giving $\mu = \frac{14}{3g} = \frac{10}{21}$ or awrt 0.48 | M1 A1 f.t. DM1 A1 | (4)
**(c)** N2L for $P$: $T - \mu 2g = 2a$ with f.t. their $\mu$, their $a$, accepting symbols, giving $T - \frac{14}{3g} \times 2g = 2 \times \frac{4}{3}$ leading to $T = 12 \text{ (N)}$ | M1 A1 f.t. DM1 A1 | (4)
**Alternatively** N2L for $Q$: $30 - T - \mu3g = 3a$ leading to $T = 12 \text{ (N)}$ | M1 A1 DM1 A1 | awrt 12
**(d)** The acceleration of $P$ and $Q$ (or the whole of the system) is the same. | B1 | (1)
**(e)** $v = u + at \Rightarrow v = \frac{4}{3} \times 3 = 4$ with N2L (for system or either particle) $-5\mu g = 5a$ giving $a = -\mu g$ and $v = u + at \Rightarrow 0 = 4 - \mu gt$ leading to $t = \frac{6}{7} \text{ (s)}$ | B1 f.t. on $a$ M1 DM1 A1 | (4) [15] Accept 0.86, 0.857
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9dbbbc01-fb66-460d-a42e-2c37ec8b451a-12_131_940_269_498}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Two particles $P$ and $Q$, of mass 2 kg and 3 kg respectively, are joined by a light inextensible string. Initially the particles are at rest on a rough horizontal plane with the string taut. A constant force $\mathbf { F }$ of magnitude 30 N is applied to $Q$ in the direction $P Q$, as shown in Figure 4. The force is applied for 3 s and during this time $Q$ travels a distance of 6 m . The coefficient of friction between each particle and the plane is $\mu$. Find
\begin{enumerate}[label=(\alph*)]
\item the acceleration of $Q$,
\item the value of $\mu$,
\item the tension in the string.
\item State how in your calculation you have used the information that the string is inextensible.
When the particles have moved for 3 s , the force $\mathbf { F }$ is removed.
\item Find the time between the instant that the force is removed and the instant that $Q$ comes to rest.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2008 Q8 [15]}}