OCR MEI M1 2016 June — Question 3 8 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Year2016
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeParticle on smooth horizontal surface, particle hanging
DifficultyModerate -0.3 This is a standard M1 pulley system question requiring straightforward application of Newton's second law to connected particles. Part (i) involves writing F=ma for each mass, part (ii) requires solving simultaneous equations (routine algebra), and part (iii) extends to an equilibrium case on an inclined plane. While it requires multiple steps and understanding of connected particles, it follows a completely standard template with no novel problem-solving or geometric insight needed—making it slightly easier than the average A-level question.
Spec3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium3.03m Equilibrium: sum of resolved forces = 0
3 Fig. 3.1 shows a block of mass 8 kg on a smooth horizontal table.
This block is connected by a light string passing over a smooth pulley to a block of mass 4 kg which hangs freely. The part of the string between the 8 kg block and the pulley is parallel to the table. The system has acceleration \(a \mathrm {~m} \mathrm {~s} ^ { - 2 }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{4c8c96cf-5184-46e4-9c45-a8a80d0a6ff8-3_330_809_525_628} \captionsetup{labelformat=empty} \caption{Fig. 3.1}
\end{figure}
  1. Write down two equations of motion, one for each block.
  2. Find the value of \(a\). The table is now tilted at an angle of \(\theta\) to the horizontal as shown in Fig. 3.2. The system is set up as before; the 4 kg block still hangs freely. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{4c8c96cf-5184-46e4-9c45-a8a80d0a6ff8-3_410_727_1324_669} \captionsetup{labelformat=empty} \caption{Fig. 3.2}
    \end{figure}
  3. The system is now in equilibrium. Find the value of \(\theta\).
(i)
AnswerMarks Guidance
AnswerMark Guidance
\(T = 8a\)B1
\(4g - T = 4a\)B1 Allow if \(a\) is in the upwards direction but the two equations must be consistent in this.
(ii)
AnswerMarks Guidance
AnswerMark Guidance
Adding the two equations \(\Rightarrow 4g = 12a\)M1 Or equivalent method. No FT from part (i).
\(a = \frac{g}{3}\) (-3.27 m s\(^{-2}\))A1 CAO but allow 3.26.
(iii)
AnswerMarks Guidance
AnswerMark Guidance
Equilibrium equations:
\(T - 4g = 0\)M1 Vertical equation
\(T - 8g \sin \theta = 0\)M1 Award if \(8g \sin \theta\) seen. Do not allow sin-cos interchange
\(4g - 8g \sin \theta = 0\)A1 Correct equation with \(T = 4g\) substituted. Note Award M1 M1 A1 for going straight to \(4g = 8g \sin \theta\) or Allow M1 M1 A0 for \(4 = 8\sin\theta\) with no previous work
\(\Rightarrow \theta = 30°\)A1 CAO
Question 3(iii) - Alternative Method
AnswerMarks Guidance
AnswerMark Guidance
\(T - 4g = 0\)M1
Triangle of forces for the 8 kg blockM1 Dependent on the other M mark. There must be an attempt to use the triangle for this mark to be awarded. The triangle must be labelled with \(4g\), \(8g\) and \(\theta\). The right angle must be drawn close to 90°.
\(\sin \theta = \frac{4g}{8g}\)A1 Dependent on both M marks.
\(\theta = 30°\)A1 CAO 
**(i)**

| Answer | Mark | Guidance |
|--------|------|----------|
| $T = 8a$ | B1 | |
| $4g - T = 4a$ | B1 | Allow if $a$ is in the upwards direction but the two equations must be consistent in this. |

**(ii)**

| Answer | Mark | Guidance |
|--------|------|----------|
| Adding the two equations $\Rightarrow 4g = 12a$ | M1 | Or equivalent method. No FT from part (i). |
| $a = \frac{g}{3}$ (-3.27 m s$^{-2}$) | A1 | CAO but allow 3.26. |

**(iii)**

| Answer | Mark | Guidance |
|--------|------|----------|
| Equilibrium equations: | | |
| $T - 4g = 0$ | M1 | Vertical equation |
| $T - 8g \sin \theta = 0$ | M1 | Award if $8g \sin \theta$ seen. Do not allow sin-cos interchange |
| $4g - 8g \sin \theta = 0$ | A1 | Correct equation with $T = 4g$ substituted. Note Award M1 M1 A1 for going straight to $4g = 8g \sin \theta$ or Allow M1 M1 A0 for $4 = 8\sin\theta$ with no previous work |
| $\Rightarrow \theta = 30°$ | A1 | CAO |

---

# Question 3(iii) - Alternative Method

| Answer | Mark | Guidance |
|--------|------|----------|
| $T - 4g = 0$ | M1 | |
| Triangle of forces for the 8 kg block | M1 | Dependent on the other M mark. There must be an attempt to use the triangle for this mark to be awarded. The triangle must be labelled with $4g$, $8g$ and $\theta$. The right angle must be drawn close to 90°. |
| $\sin \theta = \frac{4g}{8g}$ | A1 | Dependent on both M marks. |
| $\theta = 30°$ | A1 | CAO |

---
3 Fig. 3.1 shows a block of mass 8 kg on a smooth horizontal table.\\
This block is connected by a light string passing over a smooth pulley to a block of mass 4 kg which hangs freely. The part of the string between the 8 kg block and the pulley is parallel to the table.

The system has acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{4c8c96cf-5184-46e4-9c45-a8a80d0a6ff8-3_330_809_525_628}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}

(i) Write down two equations of motion, one for each block.\\
(ii) Find the value of $a$.

The table is now tilted at an angle of $\theta$ to the horizontal as shown in Fig. 3.2. The system is set up as before; the 4 kg block still hangs freely.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{4c8c96cf-5184-46e4-9c45-a8a80d0a6ff8-3_410_727_1324_669}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}

(iii) The system is now in equilibrium. Find the value of $\theta$.

\hfill \mbox{\textit{OCR MEI M1 2016 Q3 [8]}}
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