OCR MEI M1 2013 June — Question 5 7 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Year2013
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeApplied force in addition to weights
DifficultyStandard +0.3 This is a straightforward two-body mechanics problem requiring Newton's second law applied to a system and individual bodies. Students must consider two cases (acceleration left vs right) and solve linear equations, but the method is standard and well-practiced in M1 courses with no conceptual surprises.
Spec3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium

5 Fig. 5 shows blocks of mass 4 kg and 6 kg on a smooth horizontal table. They are connected by a light, inextensible string. As shown, a horizontal force \(F \mathrm {~N}\) acts on the 4 kg block and a horizontal force of 30 N acts on the 6 kg block. The magnitude of the acceleration of the system is \(2 \mathrm {~ms} ^ { - 2 }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{83e69140-4abf-4713-85da-922ce7530e47-3_106_1107_1708_479} \captionsetup{labelformat=empty} \caption{Fig. 5}
\end{figure}
  1. Find the two possible values of \(F\).
  2. Find the tension in the string in each case.

Question 5:
Part (i):
AnswerMarks Guidance
AnswerMarks Guidance
System acceleration \(= 2\) m s\(^{-2}\); net force on system \(= (4+6)\times 2 = 20\) NM1
\(30 - F = 20\) or \(F - 30 = 20\) (two cases)M1 A1
\(F = 10\) N or \(F = 50\) NA1 Both values required
Part (ii):
AnswerMarks Guidance
AnswerMarks Guidance
Case \(F=10\): acceleration is to the right. For 6 kg block: \(30 - T = 6 \times 2 \Rightarrow T = 18\) NM1 A1
Case \(F=50\): acceleration is to the left. For 6 kg block: \(T - 30 = 6 \times 2 \Rightarrow T = 42\) NA1
# Question 5:

## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| System acceleration $= 2$ m s$^{-2}$; net force on system $= (4+6)\times 2 = 20$ N | M1 | |
| $30 - F = 20$ or $F - 30 = 20$ (two cases) | M1 A1 | |
| $F = 10$ N or $F = 50$ N | A1 | Both values required |

## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Case $F=10$: acceleration is to the right. For 6 kg block: $30 - T = 6 \times 2 \Rightarrow T = 18$ N | M1 A1 | |
| Case $F=50$: acceleration is to the left. For 6 kg block: $T - 30 = 6 \times 2 \Rightarrow T = 42$ N | A1 | |

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5 Fig. 5 shows blocks of mass 4 kg and 6 kg on a smooth horizontal table. They are connected by a light, inextensible string. As shown, a horizontal force $F \mathrm {~N}$ acts on the 4 kg block and a horizontal force of 30 N acts on the 6 kg block.

The magnitude of the acceleration of the system is $2 \mathrm {~ms} ^ { - 2 }$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{83e69140-4abf-4713-85da-922ce7530e47-3_106_1107_1708_479}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}

(i) Find the two possible values of $F$.\\
(ii) Find the tension in the string in each case.

\hfill \mbox{\textit{OCR MEI M1 2013 Q5 [7]}}