CAIE M2 2018 November — Question 3 7 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2018
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeVertical elastic string: released from rest at natural length or above (string initially slack)
DifficultyStandard +0.3 This is a standard elastic string energy problem requiring conservation of energy with gravitational PE and elastic PE. Part (i) needs identifying that maximum speed occurs when forces balance (mg = λx/l), then applying energy conservation. Part (ii) is straightforward energy conservation to find maximum extension. The setup is routine for M2 level with clear methodology, though requires careful bookkeeping of energy terms across multiple steps.
Spec6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle

A particle \(P\) of mass \(0.4\text{ kg}\) is attached to a fixed point \(O\) by a light elastic string of natural length \(0.5\text{ m}\) and modulus of elasticity \(20\text{ N}\). The particle \(P\) is released from rest at \(O\).
  1. Find the greatest speed of \(P\) in the subsequent motion. [4]
  2. Find the distance below \(O\) of the point at which \(P\) comes to instantaneous rest. [3]

Question 3:

AnswerMarks Guidance
3(i)20e/0.5 = 0.4g M1
e = 0.1A1
0.4v2/2 = 0.4g(0.5 + 0.1) – 20×0.12/(2×0.5)M1 Attempt to set up a 3 term energy equation
v = 11 = 3.32A1
4

AnswerMarks Guidance
3(ii)0.4g(5 + x) = 20x2/(2×0.5) M1
[0 = 20x2 – 4x – 2] [ x = 0.432]M1 Attempt to solve a 3 term quadratic equation
Distance below O = (0.5 + 0.432) = 0.932 mA1
3
AnswerMarks Guidance
QuestionAnswer Marks
Question 3:
--- 3(i) ---
3(i) | 20e/0.5 = 0.4g | M1 | Use T = λx/L
e = 0.1 | A1
0.4v2/2 = 0.4g(0.5 + 0.1) – 20×0.12/(2×0.5) | M1 | Attempt to set up a 3 term energy equation
v = 11 = 3.32 | A1
4
--- 3(ii) ---
3(ii) | 0.4g(5 + x) = 20x2/(2×0.5) | M1 | Attempt to set up a 2 term energy equation
[0 = 20x2 – 4x – 2] [ x = 0.432] | M1 | Attempt to solve a 3 term quadratic equation
Distance below O = (0.5 + 0.432) = 0.932 m | A1
3
Question | Answer | Marks | Guidance
A particle $P$ of mass $0.4\text{ kg}$ is attached to a fixed point $O$ by a light elastic string of natural length $0.5\text{ m}$ and modulus of elasticity $20\text{ N}$. The particle $P$ is released from rest at $O$.

\begin{enumerate}[label=(\roman*)]
\item Find the greatest speed of $P$ in the subsequent motion. [4]

\item Find the distance below $O$ of the point at which $P$ comes to instantaneous rest. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2018 Q3 [7]}}