| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Two particles over pulley, vertical strings |
| Difficulty | Standard +0.2 This is a standard M1 connected particles problem requiring Newton's second law applied to a pulley system, followed by basic kinematics with constant acceleration. The 'show that' in part (i) guides students to the answer, and part (ii) is straightforward SUVAT application. Slightly easier than average due to the structured nature and standard techniques. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
\includegraphics{figure_1}
Particles $P$ and $Q$, of masses $0.3$ kg and $0.4$ kg respectively, are attached to the ends of a light inextensible string. The string passes over a smooth fixed pulley. The system is in motion with the string taut and with each of the particles moving vertically. The downward acceleration of $P$ is $a$ m s$^{-2}$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that $a = -1.4$. [4]
\end{enumerate}
Initially $P$ and $Q$ are at the same horizontal level. $P$'s initial velocity is vertically downwards and has magnitude $2.8$ m s$^{-1}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Assuming that $P$ does not reach the floor and that $Q$ does not reach the pulley, find the time taken for $P$ to return to its initial position. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 Q1 [7]}}