CAIE M2 2018 June — Question 3 5 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2018
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeNon-uniform beam on supports
DifficultyStandard +0.3 This is a standard moments equilibrium problem requiring students to take moments about the suspension point and use the center of mass of each wire segment. Part (i) involves straightforward moment calculation with basic trigonometry, while part (ii) is simple quadratic solving. The setup is clear and the method is routine for M2 students, making it slightly easier than average.
Spec6.04b Find centre of mass: using symmetry6.04e Rigid body equilibrium: coplanar forces

\includegraphics{figure_3} \(ABC\) is an object made from a uniform wire consisting of two straight portions \(AB\) and \(BC\), in which \(AB = a\), \(BC = x\) and angle \(ABC = 90°\). When the object is freely suspended from \(A\) and in equilibrium, the angle between \(AB\) and the horizontal is \(\theta\) (see diagram).
  1. Show that \(x^2 \tan \theta - 2ax - a^2 = 0\). [3]
  2. Given that \(\tan \theta = 1.25\), calculate the length of the wire in terms of \(a\). [2]

Question 3:

AnswerMarks Guidance
3(i)d = xsinθ/2 – acosθ or equivalent B1
through A
AnswerMarks Guidance
a(acosθ)/2 = x(xsinθ/2 – acosθ)M1 Take moments about A
x2tanθ – 2ax – a2 = 0 AGA1
Total:3

AnswerMarks Guidance
3(ii)1.25x2 – 2ax – a2 = 0 [x = 2a and x = – 2a/5] M1
Length ( = 2a + a ) = 3aA1
Total:2
QuestionAnswer Marks
Question 3:
--- 3(i) ---
3(i) | d = xsinθ/2 – acosθ or equivalent | B1 | Note d is the distance of the C of M of BC from the vertical
through A
a(acosθ)/2 = x(xsinθ/2 – acosθ) | M1 | Take moments about A
x2tanθ – 2ax – a2 = 0 AG | A1
Total: | 3
--- 3(ii) ---
3(ii) | 1.25x2 – 2ax – a2 = 0 [x = 2a and x = – 2a/5] | M1 | Attempts to solve the equation
Length ( = 2a + a ) = 3a | A1
Total: | 2
Question | Answer | Marks | Guidance
\includegraphics{figure_3}

$ABC$ is an object made from a uniform wire consisting of two straight portions $AB$ and $BC$, in which $AB = a$, $BC = x$ and angle $ABC = 90°$. When the object is freely suspended from $A$ and in equilibrium, the angle between $AB$ and the horizontal is $\theta$ (see diagram).

\begin{enumerate}[label=(\roman*)]
\item Show that $x^2 \tan \theta - 2ax - a^2 = 0$. [3]

\item Given that $\tan \theta = 1.25$, calculate the length of the wire in terms of $a$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2018 Q3 [5]}}