OCR Further Mechanics (Further Mechanics) 2021 June

2 The cover of a children's book is modelled as being a uniform lamina \(L . L\) occupies the region bounded by the \(x\)-axis, the curve \(y = 6 + \sin x\) and the lines \(x = 0\) and \(x = 5\) (see Fig. 2.1). The centre of mass of \(L\) is at the point \(( \bar { x } , \bar { y } )\). \begin{figure}[h] \end{figure}

1. Show that \(\bar { x } = 2.36\), correct to 3 significant figures.
2. Find \(\bar { y }\), giving your answer correct to 3 significant figures. The side of \(L\) along the \(y\)-axis is attached to the rest of the book and the book is placed on a rough horizontal plane. The attachment of the cover to the book is modelled as a hinge. The cover is held in equilibrium at an angle of \(\frac { 1 } { 3 } \pi\) radians to the horizontal by a force of magnitude \(P \mathrm {~N}\) acting at \(B\) perpendicular to the cover (see Fig. 2.2). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{d6bf2fa5-2f29-4632-b27d-ed8c5a0379cf-03_412_213_402_525} \captionsetup{labelformat=empty} \caption{Fig. 2.2}
   \end{figure}
3. State two additional modelling assumptions, one about the attachment of the cover and one about the badge, which are necessary to allow the value of \(P\) to be determined.
4. Using the modelling assumptions, determine the value of \(P\) giving your answer correct to 3 significant figures.

3 Two smooth circular discs \(A\) and \(B\) are moving on a horizontal plane. The masses of \(A\) and \(B\) are 3 kg and 4 kg respectively. At the instant before they collide

• the velocity of \(A\) is \(2 \mathrm {~ms} ^ { - 1 }\) at an angle of \(60 ^ { \circ }\) to the line joining their centres,
• the velocity of \(B\) is \(5 \mathrm {~ms} ^ { - 1 }\) towards \(A\) along the line joining their centres (see Fig. 3).

\begin{figure}[h] \end{figure} Given that the velocity of \(A\) after the collision is perpendicular to the velocity of \(A\) before the collision, find

1. the coefficient of restitution between \(A\) and \(B\),
2. the total loss of kinetic energy as a result of the collision.

4 One end of a light elastic string of natural length \(l \mathrm {~m}\) and modulus of elasticity \(\lambda \mathrm { N }\) is attached to a particle \(A\) of mass \(m \mathrm {~kg}\). The other end of the string is attached to a fixed point \(O\) which is on a horizontal surface. The surface is modelled as being smooth and \(A\) moves in a circular path around \(O\) with constant speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The extension of the string is denoted by \(x \mathrm {~m}\).

1. Show that \(x\) satisfies \(\lambda x ^ { 2 } + \lambda l x - l m v ^ { 2 } = 0\).
2. By solving the equation in part (a) and using a binomial series, show that if \(\lambda\) is very large then \(\lambda x \approx m v ^ { 2 }\).
3. By considering the tension in the string, explain how the result obtained when \(\lambda\) is very large relates to the situation when the string is inextensible. The nature of the horizontal surface is such that the modelling assumption that it is smooth is justifiable provided that the speed of the particle does not exceed \(7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). In the case where \(m = 0.16\) and \(\lambda = 260\), the extension of the string is measured as being 3.0 cm .
4. Estimate the value of \(v\).
5. Explain whether the value of \(v\) means that the modelling assumption is necessarily justifiable in this situation. \section*{Total Marks for Question Set 4: 35} \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
   1	(a)	\(\binom { 2 } { 10 } \cdot \binom { 50 } { 140 }\) \(2 \times 50 + 10 \times 140 = 1500 \mathrm {~J}\)	M1 A1 [2]	1.1 1.1	Using WD = F.x
   1	(b)	\(1500 / 5 = 300 \mathrm {~W}\)	B1ft [1]	1.1a	Their 1500	Must be a scalar value
   \multirow[t]{3}{*}{1}	\multirow[t]{3}{*}{(c)}	\(\frac { 1 } { 2 } \times 1.25 v ^ { 2 } = \frac { 1 } { 2 } \times 1.25 \times 10 ^ { 2 } + 1500\) \(50 \mathrm {~ms} ^ { - 1 }\)	M1 A1	1.1 1.1	Correct use of work-energy principle Not ±	Can use their WD for M mark
   		Alternative method \(\binom { 50 } { 140 } = 5 \mathbf { u } + \frac { 1 } { 2 } \cdot \binom { \frac { 8 } { 5 } } { 8 } \cdot 5 ^ { 2 } \Rightarrow \mathbf { u } = \binom { 6 } { 8 }\) Then \(\mathbf { v } = \binom { 6 } { 8 } + \binom { \frac { 8 } { 5 } } { 8 } .5 = \binom { 14 } { 48 }\) \(| \mathbf { v } | = 50 \mathrm {~ms} ^ { - 1 }\)	М1 A1		complete method involving constant acceleration formula(e)
   			[2]

   Question		Answer	Marks	AOs	Guidance
   2	(a)	\(\begin{aligned}	\bar { x } = \frac { \int _ { 0 } ^ { 5 } x ( 6 + \sin x ) \mathrm { d } x } { \int _ { 0 } ^ { 5 } ( 6 + \sin x ) \mathrm { d } x } = \frac { 72.622 \ldots } { 30.716 \ldots }
   	= \frac { 72.622 \ldots } { 30.716 \ldots } = 2.36 ( 3 \mathrm { sf } ) \end{aligned}\)	M1 A1 [2]	1.1 1.1	BC Attempt to use formula and either top or bottom correct soi AG. Both must be seen, or correct \(2.364 \ldots\) seen
   2	(b)	\(\bar { y } = \frac { \int _ { 0 } ^ { 5 } \frac { 1 } { 2 } ( 6 + \sin x ) ^ { 2 } \mathrm {~d} x } { \int _ { 0 } ^ { 5 } ( 6 + \sin x ) \mathrm { d } x } = \frac { 95.616 \ldots } { 30.716 \ldots }\)	M1 A1 [2]	1.1 1.1	BC Attempt to use formula and either top or bottom correct soi
   2	(c)	The (part of the) binding (attached to the cover) is light oe The CoM of the badge is at \(A\) oe	B1 B1 [2]	3.5b 3.5b	eg The binding has no mass or the binding is very small so that the mass is concentrated at the hinge or the binding is smooth eg The badge is modelled as a particle or the badge is uniform
   2	(d)	\(6 \times 2.36 \cos \frac { \pi } { 3 } + 2 \times 3 \cos \frac { \pi } { 3 }\) \(= P \times 5\) \(P = 2.02\)	M1 M1 A1 [3]	3.4 3.4 1.1	Total 'clockwise' moment about binding axis (allow inclusion of \(g\) if consistent)... ...equals 'anticlockwise' moment	May use new \(\bar { x } = 2.523 \ldots\) \(8 \cos \frac { \pi } { 3 } \times \bar { x }\) \(= 5 \mathrm { P }\)

   Question		Answer	Marks	AOs	Guidance
   \multirow[t]{3}{*}{3}	\multirow[t]{3}{*}{(a)}	\(\begin{aligned}	u _ { A y } = \sqrt { 3 } \text { or awrt } 1.73
   	v _ { A y } = u _ { A y } ( = \sqrt { 3 } )
   	\binom { 1 } { \sqrt { 3 } } \cdot \binom { v _ { A x } } { \sqrt { 3 } } = 0 \Rightarrow v _ { A x } = - 3
   	3 \times 2 \cos 60 ^ { \circ } + 4 \times - 5 = 3 \times - 3 + 4 v _ { B x }
   	v _ { B x } = - 2
   	e = \frac { - 2 - - 3 } { 2 \cos 60 ^ { \circ } - - 5 }
   	e = \frac { 1 } { 6 } \text { or awrt } 0.17 \end{aligned}\)	B1 B1 М1 М1 A1 M1 A1	3.1b 3.1b 2.2a 3.1b 1.1 3.1b 1.1	Correct perpendicular component of velocity of \(A\) before collision This component of \(A\) 's velocity is unchanged by collision Or \(\tan 30 ^ { \circ } = \frac { \sqrt { 3 } } { v _ { A x } }\). Using perpendicularity of \(A\) 's velocities to derive a value for \(v _ { A x }\) Conservation of momentum with 4 terms Restitution	or \(v _ { A } = 2 \sqrt { 3 }\) seen Could be positive if shown in diagram Allow one sign slip Could be positive Allow one sign slip
   		Alternative method for last 5 marks \(\begin{aligned} 3 \times 2 \cos 60 ^ { \circ } + 4 \times - 5 = 3 v _ { A x } + 4 v _ { B x } e = \frac { v _ { B x } - v _ { A x } } { 2 \cos 60 ^ { \circ } - - 5 } v _ { A x } = \frac { - 17 - 24 e } { 7 } \end{aligned}\) \(\binom { 1 } { \sqrt { 3 } } \cdot \binom { \frac { - 17 - 24 e } { 7 } } { \sqrt { 3 } } = 0\) \(e = \frac { 1 } { 6 }\) or awrt 0.17	М1 М1 A1 М1 A1		Conservation of momentum Restitution \(v _ { B x } = \frac { - 17 + 18 e } { 7 }\) Or \(\tan 30 ^ { \circ } = \frac { \sqrt { 3 } } { \left( \frac { 17 + 24 e } { 7 } \right) }\). Using perpendicularity of \(A\) 's velocities.	\(\begin{aligned}	3 v _ { A x } + 4 v _ { B x } = - 17
   	v _ { B x } - v _ { A x } = 6 e \end{aligned}\)
   			[7]

   Question		Answer	Marks	AOs	Guidance
   \multirow[t]{12}{*}{3}	\multirow[t]{12}{*}{(b)}		M1	3.1b	Attempt to find final speed (squared) of \(A\) vectorially	\multirow[t]{4}{*}{\(v _ { A } { } ^ { 2 } = 12\)}
   		\(\begin{aligned}	v _ { A } ^ { 2 } = ( \sqrt { 3 } ) ^ { 2 } + ( - 3 ) ^ { 2 }
   	\frac { 1 } { 2 } \times 4 \times 5 ^ { 2 } + \frac { 1 } { 2 } \times 3 \times 2 ^ { 2 }
   	\frac { 1 } { 2 } \times 4 \times 2 ^ { 2 } + \frac { 1 } { 2 } \times 3 \times 12 \end{aligned}\)	М1	1.1	Attempt to find total initial KE
   			M1	1.1	Attempt to find total final KE
   		56 or 26	A1	1.1	Either correctly calculated
   \multirow{8}{*}{}		\(56 - 26 = 30 \mathrm {~J}\)	A1	1.1	Not -30
   \multirow{7}{*}}{		Alternative method for last 4 marks		\multirow[b]{3}{*}{ } \multirow{3}{*}{} \multirow[t]{2}{*}{ М1 М1 М1 } \(\begin{aligned} \frac { 1 } { 2 } \times 4 \times 5 ^ { 2 } - \frac { 1 } { 2 } \times 4 \times 2 ^ { 2 } \frac { 1 } { 2 } \times 3 \times 2 ^ { 2 } - \frac { 1 } { 2 } \times 3 \times 12 \end{aligned}\) \(42 \text { or } 12 \text { so loss } = 42 + ( - 12 )\) Can be implied by correct \(= 30 \mathrm {~J}\) A1 Not -30 [5] Question Answer Marks AOs Guidance 4 (a) \(T = \frac { \lambda x } { l }\) and \(r = l + x\) both used in solution \(T = \frac { m v ^ { 2 } } { l + x }\) M1 М1 A1 [3] 3.3 3.3 3.3 Use of \(F = m a\) with centripetal acceleration AG 4 (b) \(\begin{aligned} x = \frac { - \lambda l + \sqrt { ( \lambda l ) ^ { 2 } - 4 \lambda \left( - l m v ^ { 2 } \right) } } { 2 \lambda } x = \frac { l } { 2 } \left( \left( 1 + \frac { 4 m v ^ { 2 } } { \lambda l } \right) ^ { \frac { 1 } { 2 } } - 1 \right) x = \frac { l } { 2 } \left( 1 + \frac { 1 } { 2 } \frac { 4 m v ^ { 2 } } { \lambda l } + \ldots - 1 \right) x \approx \frac { l } { 2 } \left( \frac { 1 } { 2 } \frac { 4 m v ^ { 2 } } { \lambda l } \right) = \frac { m v ^ { 2 } } { \lambda } \Rightarrow \lambda x \approx m v ^ { 2 } \end{aligned}\) M1 М1 A1 [3] 2.1 3.1b 1.1 Use of the quadratic equation formula Reject negative route and rearranging to form with \(\sqrt { 1 + \ldots }\) Use of binomial series AG No need to mention \(\left| \frac { 4 m v ^ { 2 } } { \lambda l } \right| < 1\) 4 (c) \(\lambda x \approx m v ^ { 2 } \Rightarrow \frac { \lambda x } { l } = T \approx \frac { m v ^ { 2 } } { l }\) and if the string were inextensible, corresponding to an infinite value of \(\lambda\) and \(x\) being 0 , then \(l\) would be the radius of the motion and so the RHS would be the centripetal force B1 [1] 3.5a 4 (d) \(v \approx \sqrt { \frac { 260 \times 0.03 } { 0.16 } } =\) awrt 7.0 B1 [1] 3.4 (6.982...) Question Answer Marks AOs Guidance 4 (e) While \(v\) in this situation is slightly below 7 nevertheless it is an estimate so we cannot be certain that the modelled value does not exceed 7 in which case the assumption would not be justified B1 2.2b