| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Three or more connected particles |
| Difficulty | Moderate -0.3 This is a straightforward two-particle pulley system requiring application of Newton's second law to two bodies with a given tension. Students must write F=ma equations for both particles, use the common acceleration, and solve algebraically. While it requires careful sign conventions and simultaneous equations, it's a standard M1 exercise with no novel insight needed—slightly easier than average due to being a direct application of a core technique. |
| Spec | 3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For \(P\): \(4mg - 3mg = 4ma\) | M1A1 | M1: Equation of motion with correct terms, condone sign errors |
| For both: \(4mg + 2mg - T = (4+2)ma = 6ma\) Any two of these | M1A1 | M1: Equation of motion with correct terms, condone sign errors |
| For \(Q\): \(2mg + 3mg - T = 2ma\) | ||
| Solve for \(T\) | DM1 | Dependent on both M's; must be in terms of \(mg\) |
| \(\frac{9mg}{2}\), \(4.5mg\) oe | A1 (6) | Any equivalent expression of the form \(kmg\) |
# Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| For $P$: $4mg - 3mg = 4ma$ | M1A1 | M1: Equation of motion with correct terms, condone sign errors |
| For both: $4mg + 2mg - T = (4+2)ma = 6ma$ **Any two of these** | M1A1 | M1: Equation of motion with correct terms, condone sign errors |
| For $Q$: $2mg + 3mg - T = 2ma$ | | |
| Solve for $T$ | DM1 | Dependent on both M's; must be in terms of $mg$ |
| $\frac{9mg}{2}$, $4.5mg$ oe | A1 (6) | Any equivalent expression of the form $kmg$ |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7a65555e-1bb2-4947-8e70-50f267017bfd-06_472_208_343_1102}
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\caption{Figure 1}
\end{center}
\end{figure}
Two particles, $P$ and $Q$, have masses $4 m$ and $2 m$ respectively. The particles are connected by a light inextensible string. A second light inextensible string has one end attached to $Q$. Both strings are taut and vertical, as shown in Figure 1.
The particles are accelerating vertically downwards.\\
Given that the tension in the string connecting the two particles is $3 m g$, find, in terms of $m$ and $g$, the tension in the upper string.
\hfill \mbox{\textit{Edexcel M1 2024 Q3 [6]}}