OCR Further Mechanics (Further Mechanics) 2021 June

1 A bungee jumper of mass 80 kg steps off a high bridge with an elastic rope attached to her ankles. She is assumed to fall vertically from rest and the air resistance she experiences is modelled as a constant force of 32 N . The rope has natural length 4 m and modulus of elasticity 470 N . By considering energy, determine the total distance she falls before first coming to instantaneous rest.

2 One end of a light inextensible string of length 0.75 m is attached to a particle \(A\) of mass 2.8 kg . The other end of the string is attached to a fixed point \(O . A\) is projected horizontally with speed \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point 0.75 m vertically above \(O\) (see Fig. 2). When \(O A\) makes an angle \(\theta\) with the upward vertical the speed of \(A\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
\(\xrightarrow [ A \text { a } ] { 6 \mathrm {~m} \mathrm {~s} ^ { - 1 } }\) Fig. 2

1. Show that \(v ^ { 2 } = 50.7 - 14.7 \cos \theta\).
2. Given that the string breaks when the tension in it reaches 200 N , find the angle that \(O A\) turns through between the instant that \(A\) is projected and the instant that the string breaks.

3 The resistive force, \(F\), on a sphere falling through a viscous fluid is thought to depend on the radius of the sphere, \(r\), the velocity of the sphere, \(v\), and the viscosity of the fluid, \(\eta\). You are given that \(\eta\) is measured in \(\mathrm { Nm } ^ { - 2 } \mathrm {~s}\).

1. By considering its units, find the dimensions of viscosity. A model of the resistive force suggests the following relationship: \(F = 6 \pi \eta ^ { \alpha } r ^ { \beta } v ^ { \gamma }\).
2. Explain whether or not it is possible to use dimensional analysis to verify that the constant \(6 \pi\) is correct.
3. Use dimensional analysis to find the values of \(\alpha , \beta\) and \(\gamma\). A sphere of radius \(r\) and mass \(m\) falls vertically from rest through the fluid. After a time \(t\) its velocity is \(v\).
4. By setting up and solving a differential equation, show that \(\mathrm { e } ^ { - k t } = \frac { g - k v } { g }\) where \(k = \frac { 6 \pi \eta r } { m }\). As the time increases, the velocity of the sphere tends towards a limit called the terminal velocity.
5. Find, in terms of \(g\) and \(k\), the terminal velocity of the sphere. In a sequence of experiments the sphere is allowed to fall through fluids of different viscosity, ranging from small to very large, with all other conditions being constant. The terminal velocity of the sphere through each fluid is measured.
6. Describe how, according to the model, the terminal velocity of the sphere changes as the viscosity of the fluid through which it falls increases.

4 Fig. 4.1 shows a uniform lamina in the shape of a sector of a circle of radius \(r\) and angle \(2 \theta\) where \(\theta\) is in radians. The sector consists of a triangle \(O A B\) and a segment bounded by the chord \(A B\). \begin{figure}[h] \end{figure}

1. Explain why the centre of mass of the segment lies on the radius through the midpoint of \(A B\).
2. Show that the distance of the centre of mass of the segment from \(O\) is \(\frac { 2 r \sin ^ { 3 } \theta } { 3 ( \theta - \sin \theta \cos \theta ) }\). A uniform circular lamina of radius 5 units is placed with its centre at the origin, \(O\), of an \(x - y\) coordinate system. A component for a machine is made by removing and discarding a segment from the lamina. The radius of the circle from which the segment is formed is 3 units and the centre of this circle is \(O\). The centre of the straight edge of the segment has coordinates \(( 0,2 )\) and this edge is perpendicular to the \(y\)-axis (see Fig. 4.2). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8859baf3-f8e8-4fbf-b54f-34f550b02c26-03_748_743_1594_251} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
   \end{figure}
3. Find the \(y\)-coordinate of the centre of mass of the component, giving your answer correct to 3 significant figures.
   \(C\) is the point on the component with coordinates \(( 0,5 )\). The component is now placed horizontally and supported only at \(O\). A particle of mass \(m \mathrm {~kg}\) is placed on the component at \(C\) and the component and particle are in equilibrium.
4. Find the mass of the component in terms of \(m\). Total Marks for Question Set 3: 40 \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
   b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
   A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
   c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
   d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
   e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.
   • When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
   • When a value is not given in the paper accept any answer that agrees with the correct value to \(\mathbf { 3 ~ s } . \mathbf { f }\). unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.
   Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
   Candidates using a value of \(9.80,9.81\) or 10 for \(g\) should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
   f Rules for replaced work and multiple attempts:
   • If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
   • If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
   • if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.
   For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
   If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
   h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Abbreviations}

   Abbreviations used in the mark scheme	Meaning
   dep*	Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark
   cao	Correct answer only
   ое	Or equivalent
   rot	Rounded or truncated
   soi	Seen or implied
   www	Without wrong working
   AG	Answer given
   awrt	Anything which rounds to
   BC	By Calculator
   DR	This question included the instruction: In this question you must show detailed reasoning.

   \end{table}

   Question		Answer	Marks	AOs	Guidance
   \multirow[t]{3}{*}{3}	\multirow[t]{3}{*}{(e)}	\(\mathrm { e } ^ { - k t } = \frac { g - k v } { g } \quad\) so \(t \rightarrow \infty , \Rightarrow v _ { T } = \frac { g } { k }\)	B1	3.4
   		Alternative method \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 0 \Rightarrow m g - 6 \pi \eta r v _ { T } = 0 \Rightarrow v _ { T } = \frac { m g } { 6 \pi \eta r } = \frac { g } { k }\)	B1
   			[1]
   3	(f)	As the viscosity increases the terminal velocity decreases... ...and as the viscosity tends to infinity the terminal velocity tends to 0	B1 B1 [2]	2.2a 2.2a

   Question			Answer	Marks	AOs	Guidance
   4	(a)		The central radius is a line of symmetry of the shape.	B1 [1]	2.4	Allow equal area each side
   4	(b)		Area/mass of sector \(\frac { 1 } { 2 } r ^ { 2 } \times 2 \theta\) and CoM of sector at \(\frac { 2 r \sin \theta } { 3 \theta }\) from centre used Area/mass of triangle \(( - ) \frac { 1 } { 2 } r ^ { 2 } \sin ( 2 \theta )\) and CoM of triangle at \(\frac { 2 } { 3 } r \cos \theta\) from centre used \(\frac { \frac { 1 } { 2 } r ^ { 2 } \times 2 \theta \times \frac { 2 r \sin \theta } { 3 \theta } - \frac { 1 } { 2 } r ^ { 2 } \sin 2 \theta \times \frac { 2 } { 3 } r \cos \theta } { \frac { 1 } { 2 } r ^ { 2 } \times 2 \theta - \frac { 1 } { 2 } r ^ { 2 } \sin 2 \theta }\)	B1 B1 M1 A1 [4]	1.2 1.2 3.1b 1.1	Must be used Using \(\bar { x } = \frac { \Sigma m _ { i } x _ { i } } { \Sigma m _ { i } }\) with sector and triangle of-ve mass, oe AG At least one intermediate step must be seen.	Must see change from double angle
   4	(c)		Sector angle is \(2 \cos ^ { - 1 } \frac { 2 } { 3 }\) or \(\theta = \cos ^ { - 1 } \frac { 2 } { 3 }\) Area/Mass of component: \(\pi \times 5 ^ { 2 } - \frac { 1 } { 2 } \times 3 ^ { 2 } ( 1.682 - \sin 1.682 )\) \(= 75.44\) \(\bar { y } = \frac { ( - 3.097 \ldots \times 2.406 \ldots ) } { 75.44 \ldots }\) \(= - 0.0988 ( 3 \mathrm { sf } )\)	B1 М1 A1 M1 A1 [5]	2.2a 1.1 1.1 3.1b 1.1	1.682... Attempting to find mass of component using 'negative' mass Their 3.097, 2.406 and 75.44	\(0.841 \ldots\) Allow use of 0.841 If \(\theta = 2 \cos ^ { - 1 } \frac { 2 } { 3 }\) used this gives 1.095... instead of 2.406...
   4	(d)		\(- 0.0988 M + 5 m = 0\) So the mass of the component is 50.6 m kg	M1 A1 [2]	1.1 1.1	May see \(M g\) and \(m g\)