CAIE M2 2016 November — Question 6 7 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2016
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeParticle in circular motion with string/rod
DifficultyChallenging +1.2 This is a multi-step energy conservation problem involving elastic potential energy, gravitational potential energy, and kinetic energy. While it requires careful bookkeeping of energy terms and understanding when the string becomes slack (tension = 0), the approach is standard for A-level mechanics: set up energy equations at different positions. The geometry is straightforward (vertical circle), and the calculations are routine once the setup is correct. Slightly above average difficulty due to the combination of elastic and gravitational energy, but follows a well-practiced method.
Spec6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.05f Vertical circle: motion including free fall

\includegraphics{figure_6} The diagram shows a smooth narrow tube formed into a fixed vertical circle with centre \(O\) and radius 0.9 m. A light elastic string with modulus of elasticity 8 N and natural length 1.2 m has one end attached to the highest point \(A\) on the inside of the tube. The other end of the string is attached to a particle \(P\) of mass 0.2 kg. The particle is released from rest at the lowest point on the inside of the tube. By considering energy, calculate
  1. the speed of \(P\) when it is at the same horizontal level as \(O\), [4]
  2. the speed of \(P\) at the instant when the string becomes slack. [3]

Question 6:

(ii) ---
6 (i)
AnswerMarks
(ii)EE = 8(0.9π – 1.2)2/(2 x 1.2)
8.83 = 0.2g x 0.9 + 0.2v2/2 +
8(0.9π/2 – 1.2)2/(2 x 1.2)
v = 8.29 ms–1
θ = 1.2/0.9 = 4/3 rad (=76.4°)
8.83 = 0.2g x 0.9 + 0.2g x 0.9cosθ
+ 0.2v2/2
AnswerMarks
v = 8.13 ms–1B1
M1
A1
A1
B1
M1
AnswerMarks
A14
3Initial EE = 8.83J
0.2 x 8.292/2 = 0.2g x 0.9cosθ
+ 0.2v2/2
Question 6:
--- 6 (i)
(ii) ---
6 (i)
(ii) | EE = 8(0.9π – 1.2)2/(2 x 1.2)
8.83 = 0.2g x 0.9 + 0.2v2/2 +
8(0.9π/2 – 1.2)2/(2 x 1.2)
v = 8.29 ms–1
θ = 1.2/0.9 = 4/3 rad (=76.4°)
8.83 = 0.2g x 0.9 + 0.2g x 0.9cosθ
+ 0.2v2/2
v = 8.13 ms–1 | B1
M1
A1
A1
B1
M1
A1 | 4
3 | Initial EE = 8.83J
0.2 x 8.292/2 = 0.2g x 0.9cosθ
+ 0.2v2/2
\includegraphics{figure_6}

The diagram shows a smooth narrow tube formed into a fixed vertical circle with centre $O$ and radius 0.9 m. A light elastic string with modulus of elasticity 8 N and natural length 1.2 m has one end attached to the highest point $A$ on the inside of the tube. The other end of the string is attached to a particle $P$ of mass 0.2 kg. The particle is released from rest at the lowest point on the inside of the tube. By considering energy, calculate

\begin{enumerate}[label=(\roman*)]
\item the speed of $P$ when it is at the same horizontal level as $O$, [4]
\item the speed of $P$ at the instant when the string becomes slack. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2016 Q6 [7]}}