OCR MEI Further Mechanics B AS 2022 June — Question 6 10 marks

Exam BoardOCR MEI
ModuleFurther Mechanics B AS (Further Mechanics B AS)
Year2022
SessionJune
Marks10
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TopicHooke's law and elastic energy
TypeParticle attached to two separate elastic strings
DifficultyStandard +0.3 This is a straightforward elastic string equilibrium problem requiring standard application of Hooke's law and EPE formula. Part (a) is a simple conceptual check (tensions differ so angles must differ), and part (b) involves direct substitution into EPE = Ξ»xΒ²/(2l) with given values. The multi-step nature and Further Mechanics context place it slightly above average, but it requires no novel insight or complex problem-solving.
Spec6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.04e Rigid body equilibrium: coplanar forces
6 Two identical light elastic strings, each of length \(l\) and modulus of elasticity \(\lambda m g\) are attached to a particle \(P\) of mass \(m\). The other end of the first string is attached to a fixed point A , and the other end of the second string is attached to a fixed point B . The points A and B are such that A is above and to the right of B and both strings are taut. The string attached to A makes an angle of \(30 ^ { \circ }\) with the vertical, and the string attached to B makes an angle of \(\theta ^ { \circ }\) with the horizontal, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{feb9a438-26b0-41d3-b044-6acd6efccde0-6_546_533_699_242} The system is in equilibrium in a vertical plane. The extension of the string attached to A is 0.9 l and the extension of the string attached to B is \(0.5 l\).
  1. Explain how you know that APB is not a straight line.
  2. Show that the elastic potential energy stored in string AP is \(k m g l\), where the value of \(k\) is to be determined correct to \(\mathbf { 3 }\) significant figures.
Question 6:
AnswerMarks Guidance
6(a) Because of the mass attached to the strings
there needs to be a force
balancing out the weight – if
straight line then no force
perpendicular to this from
the springs so nothing
balancing out the compenent
of weight in this direction
[1]
AnswerMarks
(b)πœ†π‘šπ‘”Γ—0.5𝑙 πœ†π‘šπ‘”Γ—0.9𝑙
cosπœƒ = cos60Β°
AnswerMarks
𝑙 𝑙M1*
A12.1
1.1Equate horizontal components of
tensions
T and T both seen correct in terms
A B
AnswerMarks
of Ξ», l, m, and gM1 for attempt to resolve
hoz (could be of form
T cos 60o = T cosΞΈ)
A B
AnswerMarks Guidance
(πœƒ =)25.8(Β°)A1 1.1
πœ†π‘šπ‘”Γ—0.5𝑙 πœ†π‘šπ‘”Γ—0.9𝑙
sinπœƒ+π‘šπ‘” = cos30Β°
AnswerMarks
𝑙 𝑙M1*
A13.3
1.1Resolve vertically
T and T both seen correct interms
A B
AnswerMarks
of Ξ», l, m, and gFor M1 tensions can be of
form T and T
A B
AnswerMarks Guidance
πœ†(0.9cos30Β°βˆ’0.5sinπœƒ) = 1M1* 1.1
value for ΞΈ
AnswerMarks Guidance
(πœ† =) 1.78A1 1.1
1.78
πœ†π‘šπ‘”Γ—(0.9𝑙)2
Energy =
AnswerMarks Guidance
2𝑙M1dep
*3.4 Dependent on all three previous
method marks being awarded.
AnswerMarks Guidance
0.721mglA1 2.2a
[9]
PMT
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Question 6:
6 | (a) | Because of the mass attached to the strings | B1 | 3.5b | oe | Or something stating that
there needs to be a force
balancing out the weight – if
straight line then no force
perpendicular to this from
the springs so nothing
balancing out the compenent
of weight in this direction
[1]
(b) | πœ†π‘šπ‘”Γ—0.5𝑙 πœ†π‘šπ‘”Γ—0.9𝑙
cosπœƒ = cos60Β°
𝑙 𝑙 | M1*
A1 | 2.1
1.1 | Equate horizontal components of
tensions
T and T both seen correct in terms
A B
of Ξ», l, m, and g | M1 for attempt to resolve
hoz (could be of form
T cos 60o = T cosΞΈ)
A B
(πœƒ =)25.8(Β°) | A1 | 1.1 | 25.84193… | OR: and so
πœ†π‘šπ‘”Γ—0.5𝑙 πœ†π‘šπ‘”Γ—0.9𝑙
sinπœƒ+π‘šπ‘” = cos30Β°
𝑙 𝑙 | M1*
A1 | 3.3
1.1 | Resolve vertically
T and T both seen correct interms
A B
of Ξ», l, m, and g | For M1 tensions can be of
form T and T
A B
πœ†(0.9cos30Β°βˆ’0.5sinπœƒ) = 1 | M1* | 1.1 | Attempt to find Ξ» | Must include their numerical
value for ΞΈ
(πœ† =) 1.78 | A1 | 1.1 | 1.77969… | Accept answers rounding to
1.78
πœ†π‘šπ‘”Γ—(0.9𝑙)2
Energy =
2𝑙 | M1dep
* | 3.4 | Dependent on all three previous
method marks being awarded.
0.721mgl | A1 | 2.2a
[9]
PMT
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/company/ocr
/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. Β© OCR
2022 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
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Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
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information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.
6 Two identical light elastic strings, each of length $l$ and modulus of elasticity $\lambda m g$ are attached to a particle $P$ of mass $m$.

The other end of the first string is attached to a fixed point A , and the other end of the second string is attached to a fixed point B . The points A and B are such that A is above and to the right of B and both strings are taut.

The string attached to A makes an angle of $30 ^ { \circ }$ with the vertical, and the string attached to B makes an angle of $\theta ^ { \circ }$ with the horizontal, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{feb9a438-26b0-41d3-b044-6acd6efccde0-6_546_533_699_242}

The system is in equilibrium in a vertical plane. The extension of the string attached to A is 0.9 l and the extension of the string attached to B is $0.5 l$.
\begin{enumerate}[label=(\alph*)]
\item Explain how you know that APB is not a straight line.
\item Show that the elastic potential energy stored in string AP is $k m g l$, where the value of $k$ is to be determined correct to $\mathbf { 3 }$ significant figures.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics B AS 2022 Q6 [10]}}
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