OCR MEI M1 2015 June — Question 2 8 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Year2015
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeThree or more connected particles
DifficultyStandard +0.3 This is a standard three-particle pulley system requiring force diagrams and Newton's second law applied to connected particles. While it involves multiple bodies and two tensions, the setup is straightforward (smooth surfaces, light strings), and the solution follows a routine method of writing equations for each mass and solving simultaneously. Slightly above average difficulty due to the three-body system, but still a textbook exercise with no novel insight required.
Spec3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium

2 Fig. 2 shows a 6 kg block on a smooth horizontal table. It is connected to blocks of mass 2 kg and 9 kg by two light strings which pass over smooth pulleys at the edges of the table. The parts of the strings attached to the 6 kg block are horizontal. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f87e062a-fdf2-45cf-8bc0-d05683b28e1a-2_344_1143_1352_443} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure}
  1. Draw three separate diagrams showing all the forces acting on each of the blocks.
  2. Calculate the acceleration of the system and the tension in each string.

Question 2(i):
AnswerMarks Guidance
AnswerMark Guidance
Diagrams for 2 kg and 9 kg blocks with \(T_1\), \(T_2\)B1 Diagrams for both 2 and 9 kg blocks. Tensions must be different from each other. No extra forces.
Tensions on 6 kg blockB1 Tensions must be different from each other. No extra forces.
\(6g\) and \(R\) on 6 kg blockB1 No extra forces. Special Case: when tensions given as \(T_1, T_2, T_3, T_4\) award up to SC1 SC0 for first two marks.
Question 2(ii):
AnswerMarks Guidance
AnswerMark Guidance
\(9g - T_2 = 9a\)M1 First equation correct
\(T_2 - T_1 = 6a\) and \(T_1 - 2g = 2a\)M1 Both remaining two equations correct. Do not give this mark if both tensions shown as same.
\(a = \frac{7}{17}g = 4.04 \text{ ms}^{-2}\)A1 \(a\), \(T_1\) and \(T_2\) may be found in any order; FT from first of these found
\(T_1 = 27.7\) NA1
\(T_2 = 51.9\) NA1
Alternative (Whole system):
AnswerMarks Guidance
AnswerMark Guidance
\(9g - 2g = 17a\)M1
\(a = \frac{7g}{17} = 4.04\)A1
\(T_1 - 2g = 2a\) and \(9g - T_2 = 9a\)M1 Both equations correct.
\(T_1 = 27.7\) NA1 Final two marks dependent on both M marks. \(T_1\) and \(T_2\) may be found in either order; FT from their value of \(a\).
\(T_2 = 51.9\) NA1
## Question 2(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| Diagrams for 2 kg and 9 kg blocks with $T_1$, $T_2$ | B1 | Diagrams for both 2 and 9 kg blocks. Tensions must be different from each other. No extra forces. |
| Tensions on 6 kg block | B1 | Tensions must be different from each other. No extra forces. |
| $6g$ and $R$ on 6 kg block | B1 | No extra forces. Special Case: when tensions given as $T_1, T_2, T_3, T_4$ award up to SC1 SC0 for first two marks. |

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## Question 2(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $9g - T_2 = 9a$ | M1 | First equation correct |
| $T_2 - T_1 = 6a$ and $T_1 - 2g = 2a$ | M1 | Both remaining two equations correct. Do not give this mark if both tensions shown as same. |
| $a = \frac{7}{17}g = 4.04 \text{ ms}^{-2}$ | A1 | $a$, $T_1$ and $T_2$ may be found in any order; FT from first of these found |
| $T_1 = 27.7$ N | A1 | |
| $T_2 = 51.9$ N | A1 | |

**Alternative (Whole system):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $9g - 2g = 17a$ | M1 | |
| $a = \frac{7g}{17} = 4.04$ | A1 | |
| $T_1 - 2g = 2a$ and $9g - T_2 = 9a$ | M1 | Both equations correct. |
| $T_1 = 27.7$ N | A1 | Final two marks dependent on both M marks. $T_1$ and $T_2$ may be found in either order; FT from their value of $a$. |
| $T_2 = 51.9$ N | A1 | |

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2 Fig. 2 shows a 6 kg block on a smooth horizontal table. It is connected to blocks of mass 2 kg and 9 kg by two light strings which pass over smooth pulleys at the edges of the table. The parts of the strings attached to the 6 kg block are horizontal.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f87e062a-fdf2-45cf-8bc0-d05683b28e1a-2_344_1143_1352_443}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

(i) Draw three separate diagrams showing all the forces acting on each of the blocks.\\
(ii) Calculate the acceleration of the system and the tension in each string.

\hfill \mbox{\textit{OCR MEI M1 2015 Q2 [8]}}