Edexcel M3 2012 June — Question 5 12 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2012
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeParticle on smooth curved surface
DifficultyStandard +0.8 This is a classic particle-on-sphere problem requiring energy conservation and circular motion analysis. Part (a) is straightforward application of energy methods (4 marks), but part (b) requires recognizing that loss of contact occurs when normal reaction becomes zero, setting up the equation of motion in the radial direction, and solving simultaneously with the energy equation—a multi-step problem requiring physical insight beyond routine application.
Spec6.05e Radial/tangential acceleration

A fixed smooth sphere has centre \(O\) and radius \(a\). A particle \(P\) is placed on the surface of the sphere at the point \(A\), where \(OA\) makes an angle \(\alpha\) with the upward vertical through \(O\). The particle is released from rest at \(A\). When \(OP\) makes an angle \(\theta\) to the upward vertical through \(O\), \(P\) is on the surface of the sphere and the speed of \(P\) is \(v\). Given that \(\cos \alpha = \frac{3}{5}\)
  1. show that $$v^2 = \frac{2ga}{5}(3 - 5\cos \theta)$$ [4]
  2. find the speed of \(P\) at the instant when it loses contact with the sphere. [8]

A fixed smooth sphere has centre $O$ and radius $a$. A particle $P$ is placed on the surface of the sphere at the point $A$, where $OA$ makes an angle $\alpha$ with the upward vertical through $O$. The particle is released from rest at $A$. When $OP$ makes an angle $\theta$ to the upward vertical through $O$, $P$ is on the surface of the sphere and the speed of $P$ is $v$.

Given that $\cos \alpha = \frac{3}{5}$

\begin{enumerate}[label=(\alph*)]
\item show that
$$v^2 = \frac{2ga}{5}(3 - 5\cos \theta)$$ [4]
\item find the speed of $P$ at the instant when it loses contact with the sphere. [8]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2012 Q5 [12]}}