CAIE M2 2010 November — Question 5 9 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2010
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeParticle at midpoint of string between two horizontal fixed points: vertical motion
DifficultyStandard +0.3 This is a standard two-part Hooke's law problem requiring (i) resolving forces at equilibrium using geometry and tension formula, and (ii) applying energy conservation with elastic potential energy. While it involves multiple steps and careful geometry (Pythagoras to find extensions), the techniques are routine for M2 students with no novel problem-solving required. Slightly easier than average due to the 'show that' structure in part (i) providing a target value.
Spec6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings

\includegraphics{figure_5} A light elastic string has natural length \(2\) m and modulus of elasticity \(\lambda\) N. The ends of the string are attached to fixed points \(A\) and \(B\) which are at the same horizontal level and \(2.4\) m apart. A particle \(P\) of mass \(0.6\) kg is attached to the mid-point of the string and hangs in equilibrium at a point \(0.5\) m below \(AB\) (see diagram).
  1. Show that \(\lambda = 26\). [4]
\(P\) is projected vertically downwards from the equilibrium position, and comes to instantaneous rest at a point \(0.9\) m below \(AB\).
  1. Calculate the speed of projection of \(P\). [5]

Question 5:
AnswerMarks Guidance
5(i) T = λ ( 1.22 +0.52 – 1)/1 B1
T = 0.3 or T = 0.3x26
AnswerMarks
2xTx0.5/1.3 = 6B1
λ
AnswerMarks
T = 0.3 = 7.8M1
λ
AnswerMarks
= 26 AGA1
[4]
AnswerMarks Guidance
(ii) EE1 = 2x26x0.3 2 /2x1M1 (= 2.34) Use of EPE formula, either
1.22 +0.92 2
AnswerMarks Guidance
EE2 = 2x26( – 1) /2x1A1 (= 6.5) Both expressions correct
M1Conservation of energy (including
KE/GPE/EPE)
2
AnswerMarks
0.6v /2 + 0.6x10x(0.9 – 0.5) = 6.5 – 2.34A1
–1
AnswerMarks
V = 2.42msA1
[5]
AnswerMarks Guidance
Page 6Mark Scheme: Teachers’ version Syllabus
GCE A LEVEL – October/November 20109709 53
Question 5:
5 | (i) T = λ ( 1.22 +0.52 – 1)/1 | B1 | λ
T = 0.3 or T = 0.3x26
2xTx0.5/1.3 = 6 | B1
λ
T = 0.3 = 7.8 | M1
λ
= 26 AG | A1
[4]
(ii) EE1 = 2x26x0.3 2 /2x1 | M1 | (= 2.34) Use of EPE formula, either
1.22 +0.92 2
EE2 = 2x26( – 1) /2x1 | A1 | (= 6.5) Both expressions correct
M1 | Conservation of energy (including
KE/GPE/EPE)
2
0.6v /2 + 0.6x10x(0.9 – 0.5) = 6.5 – 2.34 | A1
–1
V = 2.42ms | A1
[5]
Page 6 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – October/November 2010 | 9709 | 53
\includegraphics{figure_5}

A light elastic string has natural length $2$ m and modulus of elasticity $\lambda$ N. The ends of the string are attached to fixed points $A$ and $B$ which are at the same horizontal level and $2.4$ m apart. A particle $P$ of mass $0.6$ kg is attached to the mid-point of the string and hangs in equilibrium at a point $0.5$ m below $AB$ (see diagram).

\begin{enumerate}[label=(\roman*)]
\item Show that $\lambda = 26$. [4]
\end{enumerate}

$P$ is projected vertically downwards from the equilibrium position, and comes to instantaneous rest at a point $0.9$ m below $AB$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Calculate the speed of projection of $P$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2010 Q5 [9]}}