OCR M3 2007 January — Question 7 13 marks

Exam BoardOCR
ModuleM3 (Mechanics 3)
Year2007
SessionJanuary
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNewton-Raphson method
TypeApplied context requiring Newton-Raphson
DifficultyChallenging +1.2 This is a multi-step mechanics problem combining elastic strings, circular motion, and numerical methods. Part (i)(a) requires force resolution to derive an equation (standard A-level mechanics), part (i)(b) applies a given iterative formula (routine calculation), and part (ii) uses energy conservation with some geometric reasoning. While it requires careful setup and multiple techniques, each individual step follows standard A-level patterns without requiring novel insight. The Newton-Raphson connection is simplified to a given iterative formula, making it more accessible than deriving the method from scratch.
Spec1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle

7 \includegraphics[max width=\textwidth, alt={}, center]{f334f6e4-2a60-4647-8b37-e48937c85747-4_721_691_269_726} The diagram shows a particle \(P\) of mass 0.5 kg attached to the highest point \(A\) of a fixed smooth sphere by a light elastic string. The sphere has centre \(O\) and radius 1.2 m . The string has natural length 0.6 m and modulus of elasticity \(6.86 \mathrm {~N} . P\) is released from rest at a point on the surface of the sphere where the acute angle \(A O P\) is at least 0.5 radians.
  1. (a) For the case angle \(A O P = \alpha , P\) remains at rest. Show that \(\sin \alpha = 2.8 \alpha - 1.4\).
    (b) Use the iterative formula $$\alpha _ { n + 1 } = \frac { \sin \alpha _ { n } } { 2.8 } + 0.5 ,$$ with \(\alpha _ { 1 } = 0.8\), to find \(\alpha\) correct to 2 significant figures.
  2. Given instead that angle \(A O P = 0.5\) radians when \(P\) is released, find the speed of \(P\) when angle \(A O P = 0.8\) radians, given that \(P\) is at all times in contact with the surface of the sphere. State whether the speed of \(P\) is increasing or decreasing when angle \(A O P = 0.8\) radians.

Question 7:
AnswerMarks Guidance
Answer/WorkingMark Guidance
(i)(a) Extension \(= 1.2\alpha - 0.6\)B1
\([T = mg\sin\alpha]\)M1 For resolving forces tangentially
\(0.5\times9.8\sin\alpha = 6.86(1.2\alpha - 0.6)/0.6\)A1ft
\(\sin\alpha = 2.8\alpha - 1.4\)A1 4 AG
(i)(b) \([0.8, 0.756\ldots, 0.745\ldots, 0.742\ldots, 0.741\ldots, 0.741\ldots]\)M1 For attempting to find \(\alpha_2\) and \(\alpha_3\)
\(\alpha = 0.74\)A1 2
(ii) \(\Delta h = 1.2(\cos0.5 - \cos0.8)\) \([0.217\ldots]\)B1
\([0.5\times9.8\times0.217\ldots = 1.06355\ldots]\)M1 For using \(\Delta(\text{PE}) = mg\Delta h\)
\([6.86(1.2\times0.8-0.6)^2/(2\times0.6) = 0.74088]\)M1 For using \(\text{EE} = \lambda x^2/2L\)
M1For using the principle of conservation of energy
\(\frac{1}{2}(0.5)v^2 = 1.06355\ldots - 0.74088\)A1 Any correct equation for \(v^2\)
Speed is \(1.14\text{ms}^{-1}\)A1
Speed is decreasingB1ft 7
# Question 7:

| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)(a)** Extension $= 1.2\alpha - 0.6$ | B1 | |
| $[T = mg\sin\alpha]$ | M1 | For resolving forces tangentially |
| $0.5\times9.8\sin\alpha = 6.86(1.2\alpha - 0.6)/0.6$ | A1ft | |
| $\sin\alpha = 2.8\alpha - 1.4$ | A1 | **4** AG |
| **(i)(b)** $[0.8, 0.756\ldots, 0.745\ldots, 0.742\ldots, 0.741\ldots, 0.741\ldots]$ | M1 | For attempting to find $\alpha_2$ and $\alpha_3$ |
| $\alpha = 0.74$ | A1 | **2** |
| **(ii)** $\Delta h = 1.2(\cos0.5 - \cos0.8)$ $[0.217\ldots]$ | B1 | |
| $[0.5\times9.8\times0.217\ldots = 1.06355\ldots]$ | M1 | For using $\Delta(\text{PE}) = mg\Delta h$ |
| $[6.86(1.2\times0.8-0.6)^2/(2\times0.6) = 0.74088]$ | M1 | For using $\text{EE} = \lambda x^2/2L$ |
| | M1 | For using the principle of conservation of energy |
| $\frac{1}{2}(0.5)v^2 = 1.06355\ldots - 0.74088$ | A1 | Any correct equation for $v^2$ |
| Speed is $1.14\text{ms}^{-1}$ | A1 | |
| Speed is decreasing | B1ft | **7** |
7\\
\includegraphics[max width=\textwidth, alt={}, center]{f334f6e4-2a60-4647-8b37-e48937c85747-4_721_691_269_726}

The diagram shows a particle $P$ of mass 0.5 kg attached to the highest point $A$ of a fixed smooth sphere by a light elastic string. The sphere has centre $O$ and radius 1.2 m . The string has natural length 0.6 m and modulus of elasticity $6.86 \mathrm {~N} . P$ is released from rest at a point on the surface of the sphere where the acute angle $A O P$ is at least 0.5 radians.
\begin{enumerate}[label=(\roman*)]
\item (a) For the case angle $A O P = \alpha , P$ remains at rest. Show that $\sin \alpha = 2.8 \alpha - 1.4$.\\
(b) Use the iterative formula

$$\alpha _ { n + 1 } = \frac { \sin \alpha _ { n } } { 2.8 } + 0.5 ,$$

with $\alpha _ { 1 } = 0.8$, to find $\alpha$ correct to 2 significant figures.
\item Given instead that angle $A O P = 0.5$ radians when $P$ is released, find the speed of $P$ when angle $A O P = 0.8$ radians, given that $P$ is at all times in contact with the surface of the sphere. State whether the speed of $P$ is increasing or decreasing when angle $A O P = 0.8$ radians.
\end{enumerate}

\hfill \mbox{\textit{OCR M3 2007 Q7 [13]}}