CAIE M2 2018 June — Question 5 8 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2018
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeElastic string on smooth inclined plane
DifficultyChallenging +1.2 This is a multi-step energy problem requiring Hooke's law, elastic EPE formula, and energy conservation. Part (i) is straightforward application of EPE = λx²/2l. Part (ii) requires identifying that maximum speed occurs at natural length, then applying energy conservation across the motion. While it involves several concepts and careful bookkeeping of energy terms, the approach is standard for M2 elastic string problems with no novel geometric or conceptual insights required.
Spec6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings

A particle \(P\) of mass \(0.7 \text{ kg}\) is attached by a light elastic string to a fixed point \(O\) on a smooth plane inclined at an angle of \(30°\) to the horizontal. The natural length of the string is \(0.5 \text{ m}\) and the modulus of elasticity is \(20 \text{ N}\). The particle \(P\) is projected up the line of greatest slope through \(O\) from a point \(A\) below the level of \(O\). The initial kinetic energy of \(P\) is \(1.8 \text{ J}\) and the initial elastic potential energy in the string is also \(1.8 \text{ J}\).
  1. Find the distance \(OA\). [2]
  2. Find the greatest speed of \(P\) in the motion. [6]

Question 5:

AnswerMarks
5(i)20e2
1.8=
AnswerMarks Guidance
( 2×0.5 )M1 λx
Use T =
l
AnswerMarks
e = 0.3, OA = 0.8A1
2

AnswerMarks
5(ii)20x
0.7gsin30 =
AnswerMarks Guidance
0.5M1 Use Newton’s Second Law
up the plane
AnswerMarks
x = 0.0875 mA1
20×0.08752
EPE=
AnswerMarks
( 2×0.5 )B1
0.7v2 20×0.08752
=1.8+1.8−0.7g ( 0.3−0.0875 ) sin30−
AnswerMarks Guidance
2 ( 2×0.5 )M1 Attempt to set up a 5 term
energy equation
AnswerMarks
A1Correct equation
v = 2.78 ms−1A1
6
AnswerMarks Guidance
QuestionAnswer Marks
Question 5:
--- 5(i) ---
5(i) | 20e2
1.8=
( 2×0.5 ) | M1 | λx
Use T =
l
e = 0.3, OA = 0.8 | A1
2
--- 5(ii) ---
5(ii) | 20x
0.7gsin30 =
0.5 | M1 | Use Newton’s Second Law
up the plane
x = 0.0875 m | A1
20×0.08752
EPE=
( 2×0.5 ) | B1
0.7v2 20×0.08752
=1.8+1.8−0.7g ( 0.3−0.0875 ) sin30−
2 ( 2×0.5 ) | M1 | Attempt to set up a 5 term
energy equation
A1 | Correct equation
v = 2.78 ms−1 | A1
6
Question | Answer | Marks | Guidance
A particle $P$ of mass $0.7 \text{ kg}$ is attached by a light elastic string to a fixed point $O$ on a smooth plane inclined at an angle of $30°$ to the horizontal. The natural length of the string is $0.5 \text{ m}$ and the modulus of elasticity is $20 \text{ N}$. The particle $P$ is projected up the line of greatest slope through $O$ from a point $A$ below the level of $O$. The initial kinetic energy of $P$ is $1.8 \text{ J}$ and the initial elastic potential energy in the string is also $1.8 \text{ J}$.

\begin{enumerate}[label=(\roman*)]
\item Find the distance $OA$. [2]
\item Find the greatest speed of $P$ in the motion. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2018 Q5 [8]}}