OCR MEI M1 — Question 1 8 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNewton's laws and connected particles
TypeLift with passenger or load
DifficultyStandard +0.3 This is a straightforward application of Newton's second law with standard lift mechanics. Parts (i) and (ii) are routine F=ma calculations with vertical forces only. Part (iii) adds a simple angle component requiring resolution of forces, but remains a standard textbook exercise requiring no novel insight—slightly easier than average due to its methodical, step-by-step nature.
Spec3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03f Weight: W=mg

1 Fig. 1.1 shows a circular cylinder of mass 100 kg being raised by a light, inextensible vertical wire AB . There is negligible air resistance. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{bf477f61-9f8f-418a-86d8-392bc30323b1-1_306_256_368_965} \captionsetup{labelformat=empty} \caption{Fig. 1.1}
\end{figure}
  1. Calculate the acceleration of the cylinder when the tension in the wire is 1000 N .
  2. Calculate the tension in the wire when the cylinder has an upward acceleration of \(0.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). The cylinder is now raised inside a fixed smooth vertical tube that prevents horizontal motion but provides negligible resistance to the upward motion of the cylinder. When the wire is inclined at \(30 ^ { \circ }\) to the vertical, as shown in Fig. 1.2, the cylinder again has an upward acceleration of \(0.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{bf477f61-9f8f-418a-86d8-392bc30323b1-1_308_490_1230_849} \captionsetup{labelformat=empty} \caption{Fig. 1.2}
    \end{figure}
  3. Calculate the new tension in the wire.

1 Fig. 1.1 shows a circular cylinder of mass 100 kg being raised by a light, inextensible vertical wire AB . There is negligible air resistance.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{bf477f61-9f8f-418a-86d8-392bc30323b1-1_306_256_368_965}
\captionsetup{labelformat=empty}
\caption{Fig. 1.1}
\end{center}
\end{figure}

(i) Calculate the acceleration of the cylinder when the tension in the wire is 1000 N .\\
(ii) Calculate the tension in the wire when the cylinder has an upward acceleration of $0.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.

The cylinder is now raised inside a fixed smooth vertical tube that prevents horizontal motion but provides negligible resistance to the upward motion of the cylinder. When the wire is inclined at $30 ^ { \circ }$ to the vertical, as shown in Fig. 1.2, the cylinder again has an upward acceleration of $0.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{bf477f61-9f8f-418a-86d8-392bc30323b1-1_308_490_1230_849}
\captionsetup{labelformat=empty}
\caption{Fig. 1.2}
\end{center}
\end{figure}

(iii) Calculate the new tension in the wire.

\hfill \mbox{\textit{OCR MEI M1  Q1 [8]}}