4 Particles \(A , B\) and \(C\) of masses \(2 \mathrm {~kg} , 3 \mathrm {~kg}\) and 5 kg respectively are joined by light rigid rods to form a triangular frame. The frame is placed at rest on a horizontal plane with \(A\) at the point \(( 0,0 )\), \(B\) at the point ( \(0.6,0\) ) and \(C\) at the point ( \(0.4,0.2\) ), where distances in the coordinate system are measured in metres (see Fig. 1).
\begin{figure}[h]
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\caption{Fig. 1}
\end{figure}
\(G\), which is the centre of mass of the frame, is at the point \(( \bar { x } , \bar { y } )\).
- - Show that \(\bar { x } = 0.38\).
- Find \(\bar { y }\).
- Explain why it would be impossible for the frame to be in equilibrium in a horizontal plane supported at only one point.
A rough plane, \(\Pi\), is inclined at an angle \(\theta\) to the horizontal where \(\sin \theta = \frac { 3 } { 5 }\). The frame is placed on \(\Pi\) with \(A B\) vertical and \(B\) in contact with \(\Pi . C\) is in the same vertical plane as \(A B\) and a line of greatest slope of \(\Pi . C\) is on the down-slope side of \(A B\).
The frame is kept in equilibrium by a horizontal light elastic string whose natural length is \(l \mathrm {~m}\) and whose modulus of elasticity is \(g \mathrm {~N}\). The string is attached to \(A\) at one end and to a fixed point on \(\Pi\) at the other end (see Fig. 2).
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{709f3a7a-d857-4813-98ab-de6b41a3a8dc-03_605_828_1525_248}
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\caption{Fig. 2}
\end{figure}
The coefficient of friction between \(B\) and \(\Pi\) is \(\mu\). - Show that \(l = 0.3\).
- Show that \(\mu \geqslant \frac { 14 } { 27 }\).
\section*{Total Marks for Question Set 6: 38}
\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. |
| Question | Answer | Marks | AOs | Guidance |
| 1 | (a) | \(\begin{aligned} | \mathrm { KE } = \frac { 1 } { 2 } \times 2 \times \binom { - 19.5 } { - 60 } \cdot \binom { - 19.5 } { - 60 } = 19.5 ^ { 2 } + 60 ^ { 2 } |
| \text { awrt } 3980 \mathrm {~J} \end{aligned}\) | M1 \(\begin{aligned} | \text { A1 } |
| { [ 2 ] } |
| \end{aligned}\) | | Using KE \(= \frac { 1 } { 2 } m \mathbf { v } . \mathbf { v }\) or \(\frac { 1 } { 2 } m v ^ { 2 }\) and attempting to find dot product or magnitude | |
| 1 | (b) | | \(\begin{aligned} | \binom { 4 t } { - 2 } = 2 \mathbf { a } = 2 \frac { \mathrm {~d} \mathbf { v } } { \mathrm {~d} t } | | \Rightarrow \mathbf { v } = \binom { t ^ { 2 } } { - t } + \mathbf { u } \end{aligned}\) | | \(t = 0 \Rightarrow \mathbf { u } = \binom { - 19.5 } { - 60 }\) so \(\mathbf { v } = \binom { t ^ { 2 } - 19.5 } { - t - 60 }\) oe \(P = \mathbf { F } \cdot \mathbf { v } = \binom { 4 t } { - 2 } \cdot \binom { t ^ { 2 } - 19.5 } { - t - 60 } = 4 t ^ { 3 } - 76 t + 120\) | | \(4 t ^ { 3 } - 76 t + 120 = 0\) | | \(t = 2,3\) |
| | | | Using \(\mathbf { F } = m \mathbf { a }\) with \(\mathbf { a } = \frac { \mathrm { d } \mathbf { v } } { \mathrm { d } t }\) soi | | Integrating to find \(\mathbf { v }\) (u may be missing or may look like a scalar at this stage) | | Using \(P =\) F.v and attempt at dot product | | \(\mathbf { B C }\) If \(t = - 5\) included then A0 |
| Must be reasonable attempt at integration |
| Question | Answer | Marks | AOs | Guidance |
| \multirow[t]{7}{*}{2} | | | M1 | 1.1 | Conservation of momentum | \multirow{3}{*}{} |
| | | A1 | 1.1 | Both | |
| | | \(A\) colliding with \(B\) : \(2 \times 5 u = 2 v _ { A } + 3 v _ { B }\) \(\frac { v _ { B } - v _ { A } } { 5 u } = 1\) | | \(v _ { A } = - u\) and \(v _ { B } = 4 u\) |
| M1 | 1.1 | Conservation of momentum | |
| \multirow{4}{*}{} | \(V _ { B } = \frac { u ( 3 - 5 e ) } { 2 }\) | A1 | 1.1 | Correct \(V _ { B }\) | \(V _ { C } = \frac { 3 u ( 1 + e ) } { 2 }\), but this need not be stated; ignore any wrong value seen, provided not used further |
| | \(e \leq 1 \Rightarrow V _ { B } \geq \frac { u ( 3 - 5 ) } { 2 } = - u\) | M1 | 2.1 | oe; eg stating that the greatest leftwards speed of \(B\) occurs when \(e = 1\) and evaluating this | Or could see eg assumption \(V _ { B } < v _ { A }\) leading to \(e > 1\) stated as a contradiction |
| | So \(A\) and \(B\) are both travelling in the negative direction, but \(\left| v _ { A } \right| \geq \left| v _ { B } \right|\) hence \(B\) and \(A\) do not collide again, ie there are only two collisions (AG) | E1 | 2.2a | Conclusion correctly drawn | Allow statement such as ' \(B\) is not catching up with \(A\) ' in place of a formal inequality, provided correct working is seen |
| | | [8] | | | |
| Question | Answer | Marks | AOs | Guidance |
| 3 | (a) | \(\begin{aligned} | F = k \sqrt { 9 + 1.25 ^ { 2 } } = 13 |
| k = 4 |
| 4 \sqrt { 9 + v ^ { 2 } } = 8 \frac { \mathrm {~d} v } { \mathrm {~d} t } \Rightarrow \frac { 1 } { \sqrt { 9 + v ^ { 2 } } } \frac { \mathrm {~d} v } { \mathrm {~d} t } = \frac { 1 } { 2 } \quad ( \mathbf { A G } ) \end{aligned}\) | | | | Substituting \(v = 1.25\) and \(F = 13\) | | Clear use of \(F = m a\) leading to AG |
| |
| 3 | (b) | \(\begin{aligned} | \int \frac { 1 } { \sqrt { 9 + v ^ { 2 } } } \mathrm {~d} v = \frac { 1 } { 2 } t + c |
| \sinh ^ { - 1 } \left( \frac { v } { 3 } \right) = \frac { 1 } { 2 } t + c |
| c = 0 |
| v = 3 \sinh \left( \frac { 1 } { 2 } t \right) \end{aligned}\) | | | | Separating the variables and correctly integrating to obtain bt | | Integrating other side to obtain \(p \sinh ^ { - 1 } q v\) | | Substituting \(t = 0 , v = 0\) into a solution of the DE to obtain \(c\) |
| | Condone omission of \(c\) for these two M marks | | Or using 0 as limits on integrals |
|
| 3 | (c) | \(\begin{aligned} | \frac { \mathrm { d } x } { \mathrm {~d} t } = 3 \sinh \left( \frac { 1 } { 2 } t \right) \Rightarrow x = C + 6 \cosh \left( \frac { 1 } { 2 } t \right) |
| t = 0 , x = 0 \Rightarrow C = - 6 \Rightarrow x = 6 \left( \cosh \left( \frac { 1 } { 2 } t \right) - 1 \right) \text { oe } \end{aligned}\) | | | Replacing \(v\) with \(\frac { \mathrm { d } x } { \mathrm {~d} t }\) and integrating to obtain \(r \cosh s t\) | Condone omission of \(C\) (or use of ' \(c\) ' again) |
| Question | Answer | Marks | AOs | Guidance |
| 4 | (a) | \(\begin{aligned} | \text { Use of } \bar { x } = \frac { \Sigma m x } { \Sigma m } \text { or } \bar { y } = \frac { \Sigma m y } { \Sigma m } |
| \bar { x } = \frac { 2 \times 0 + 3 \times 0.6 + 5 \times 0.4 } { 2 + 3 + 5 } = \frac { 3.8 } { 10 } = 0.38 |
| \bar { y } = \frac { 2 \times 0 + 3 \times 0 + 5 \times 0.2 } { 2 + 3 + 5 } = \frac { 1 } { 10 } = 0.1 \end{aligned}\) | | | | Values must be substituted | | Intermediate step must be seen |
| |
| 4 | (b) | \(G\) is not on the frame so the weight would have an unbalanced moment about the support point | | 2.4 | | |
| 4 | (c) | | Moments about \(B : 0.6 T = 10 g \times 0.1\) \(T = \frac { g ( 0.8 - l ) } { l }\) | | Eliminate \(T : \frac { 5 g } { 3 } = \frac { g ( 0.8 - l ) } { l } \Rightarrow 8 l = 2.4 \Rightarrow l = 0.3\) (AG) |
| | | | Each term must be of the form \(F \times d\) | | Correct form for Hooke's law | | Intermediate working must be seen |
| |
| 4 | (d) | | Any two resolving equations oe, eg: (vertically): \(R \cos \theta + F \sin \theta = 10 g\) | | (horizontally): \(R \sin \theta = F \cos \theta + T\) (parallel to plane): \(F + T \cos \theta = 10 g \sin \theta\) (perpendicular to plane): \(R = T \sin \theta + 10 g \cos \theta\) | | Use \(T = \frac { 5 } { 3 } g , \sin \theta = \frac { 3 } { 5 } , \cos \theta = \frac { 4 } { 5 }\) to derive equations \(4 R + 3 F = 50 g\) and \(9 R = 12 F + 25 g\) oe \(R = 9 g\) and \(F = \frac { 14 } { 3 } g\) | | \(F \leq \mu R \Rightarrow \frac { 14 } { 3 } g \leq \mu \times 9 g\) | | \(\therefore \mu \geq \frac { 14 g } { 3 \times 9 g } = \frac { 14 } { 27 } \quad\) (AG) |
| | | | Or taking moments eg about the point where the string is attached to the plane \(( 1 \times N = ( 0.8 + 0.1 ) \times 10 g )\) or the point where the lines of action of \(R\) and the weight intersect: \(\frac { 0.1 } { \sin \theta } \times F = \left( 0.6 - \frac { 0.1 } { \tan \theta } \right) \times T\) | | Forming equations that lead to numerical values for \(F\) and \(R\) | | Both | | \(F \leq \mu R\) with values substituted in | | Intermediate working must be seen |
| \(R\) is the normal contact force, \(F\) is the frictional force (up the slope) |