OCR MEI M4 — Question 2 12 marks

Exam BoardOCR MEI
ModuleM4 (Mechanics 4)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypePotential energy with inextensible strings or gravity only
DifficultyChallenging +1.8 This is a sophisticated M4 mechanics problem requiring energy methods for equilibrium analysis. Students must establish geometric relationships using the constraint of fixed string length, derive potential energy as a function of angle, then use calculus (dV/dθ = 0 and d²V/dθ²) to find and classify equilibrium positions. The multi-step nature, geometric insight needed, and stability analysis place it well above average difficulty, though the calculus itself is standard once the setup is complete.
Spec3.04b Equilibrium: zero resultant moment and force6.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces
2 A rigid circular hoop of radius \(a\) is fixed in a vertical plane. At the highest point of the hoop there is a small smooth pulley, P. A light inextensible string AB of length \(\frac { 5 } { 2 } a\) is passed over the pulley. A particle of mass \(m\) is attached to the string at \(\mathrm { B } . \mathrm { PB }\) is vertical and angle \(\mathrm { APB } = \theta\). A small smooth ring of mass \(m\) is threaded onto the hoop and attached to the string at A . This situation is shown in Fig. 2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{5bed3ad4-0e20-4458-a37f-655faf84c31a-02_568_549_1306_758} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure}
  1. Show that \(\mathrm { PB } = \frac { 5 } { 2 } a - 2 a \cos \theta\) and hence show that the potential energy of the system relative to P is \(V = - m g a \left( 2 \cos ^ { 2 } \theta - 2 \cos \theta + \frac { 5 } { 2 } \right)\).
  2. Hence find the positions of equilibrium and investigate their stability.
(i) Show \(v = \frac{m_0 u}{m_0 + kt}\) and find \(x\) [9]
M1: Apply conservation of momentum to coal falling vertically into truck: \(m_0 v = (m_0 + kt)v\)
or use impulse-momentum: \((m_0 + kt)\frac{dv}{dt} = 0\) (no external horizontal forces)
A1: \(m_0 u = (m_0 + kt)v\)
A1: \(v = \frac{m_0 u}{m_0 + kt}\)
M1: \(x = \int_0^t v \, dt = \int_0^t \frac{m_0 u}{m_0 + kt} \, dt\)
M1: Let \(u_1 = m_0 + kt\), then \(du_1 = k \, dt\)
A1: \(x = \frac{m_0 u}{k} \ln\left(\frac{m_0 + kt}{m_0}\right)\)
or \(x = \frac{m_0 u}{k}\left[\ln(m_0 + kt) - \ln m_0\right]\)
(ii) Distance when speed is halved [3]
M1: When \(v = \frac{u}{2}\): \(\frac{m_0 u}{m_0 + kt} = \frac{u}{2}\)
A1: \(m_0 + kt = 2m_0\), so \(t = \frac{m_0}{k}\)
A1: \(x = \frac{m_0 u}{k}\ln 2\)
**(i) Show $v = \frac{m_0 u}{m_0 + kt}$ and find $x$ [9]**

M1: Apply conservation of momentum to coal falling vertically into truck: $m_0 v = (m_0 + kt)v$

or use impulse-momentum: $(m_0 + kt)\frac{dv}{dt} = 0$ (no external horizontal forces)

A1: $m_0 u = (m_0 + kt)v$

A1: $v = \frac{m_0 u}{m_0 + kt}$

M1: $x = \int_0^t v \, dt = \int_0^t \frac{m_0 u}{m_0 + kt} \, dt$

M1: Let $u_1 = m_0 + kt$, then $du_1 = k \, dt$

A1: $x = \frac{m_0 u}{k} \ln\left(\frac{m_0 + kt}{m_0}\right)$

or $x = \frac{m_0 u}{k}\left[\ln(m_0 + kt) - \ln m_0\right]$

**(ii) Distance when speed is halved [3]**

M1: When $v = \frac{u}{2}$: $\frac{m_0 u}{m_0 + kt} = \frac{u}{2}$

A1: $m_0 + kt = 2m_0$, so $t = \frac{m_0}{k}$

A1: $x = \frac{m_0 u}{k}\ln 2$
2 A rigid circular hoop of radius $a$ is fixed in a vertical plane. At the highest point of the hoop there is a small smooth pulley, P. A light inextensible string AB of length $\frac { 5 } { 2 } a$ is passed over the pulley.

A particle of mass $m$ is attached to the string at $\mathrm { B } . \mathrm { PB }$ is vertical and angle $\mathrm { APB } = \theta$. A small smooth ring of mass $m$ is threaded onto the hoop and attached to the string at A . This situation is shown in Fig. 2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{5bed3ad4-0e20-4458-a37f-655faf84c31a-02_568_549_1306_758}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

(i) Show that $\mathrm { PB } = \frac { 5 } { 2 } a - 2 a \cos \theta$ and hence show that the potential energy of the system relative to P is $V = - m g a \left( 2 \cos ^ { 2 } \theta - 2 \cos \theta + \frac { 5 } { 2 } \right)$.\\
(ii) Hence find the positions of equilibrium and investigate their stability.

\hfill \mbox{\textit{OCR MEI M4  Q2 [12]}}
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